ON SOME CLASSES OF SEMI-DISCRETE DARBOUX INTEGRABLEEQUATIONS

A THESIS SUBMITTED TOTHE GRADUATE SCHOOL OF NATURAL AND APPLIED SCIENCES

OFMIDDLE EAST TECHNICAL UNIVERSITY

BY

ERGÜN BILEN

IN PARTIAL FULFILLMENT OF THE REQUIREMENTSFOR

THE DEGREE OF DOCTOR OF PHILOSOPHYIN

MATHEMATICS

DECEMBER 2017

Approval of the thesis:

ON SOME CLASSES OF SEMI-DISCRETE DARBOUX INTEGRABLEEQUATIONS

submitted by ERGÜN BILEN in partial fulfillment of the requirements for the degreeof Doctor of Philosophy in Mathematics Department, Middle East TechnicalUniversity by,

Prof. Dr. Gülbin Dural ÜnverDean, Graduate School of Natural and Applied Sciences

Prof. Dr. Mustafa KorkmazHead of Department, Mathematics

Assoc. Prof. Dr. Kostyantyn ZheltukhinSupervisor, Mathematics Department, METU

Examining Committee Members:

Prof. Dr. Atalay KarasuMathematics Department, METU

Assoc. Prof. Dr. Kostyantyn ZheltukhinMathematics Department, METU

Prof. Dr. Hursit ÖnsiperPhysics Department, METU

Assoc. Prof. Dr. Aslı PekcanMathematics Department, Hacettepe University

Assoc. Prof. Dr. Mehmet Onur FenMathematics Department, TED University

Date:

I hereby declare that all information in this document has been obtained andpresented in accordance with academic rules and ethical conduct. I also declarethat, as required by these rules and conduct, I have fully cited and referenced allmaterial and results that are not original to this work.

Name, Last Name: ERGÜN BILEN

Signature :

iv

ABSTRACT

ON SOME CLASSES OF SEMI-DISCRETE DARBOUX INTEGRABLEEQUATIONS

Bılen, Ergün

Ph.D., Department of Mathematics

Supervisor : Assoc. Prof. Dr. Kostyantyn Zheltukhin

December 2017, 71 pages

In this thesis we consider Darboux integrable semi-discrete hyperbolic equations ofthe form

t1x = f(t, t1, tx),∂f

∂tx6= 0.

We use the notion of characteristic Lie ring for a classification problem based ondimensions of characteristic x- and n-rings.Let A = (aij)N×N be a N × N matrix. We also consider semi-discrete hyperbolicequations of exponential type

ui1,x − uix = e∑a+iju

j1+

∑a−iju

j

i, j = 1, 2, . . . , N.

We find the conditions on aij’s so that the above equation is Darboux integrable whenN = 2.

Keywords: Darboux integrability, Characteristic Lie ring, Hyperbolic equations

v

ÖZ

YARI AYRIK DARBOUX INTEGRALLENEBILIR DENKLEMLERIN BAZIALTSINIFLARI ÜZERINE

Bılen, Ergün

Doktora, Matematik Bölümü

Tez Yöneticisi : Doç. Dr. Kostyantyn Zheltukhin

Aralık 2017 , 71 sayfa

Bu tezde,

t1x = f(t, t1, tx),∂f

∂tx6= 0,

formuna sahip Darboux integrallenebilir yarı ayrık hiperbolik denklemleri ele aldık.Karakteristik Lie halkası kavramını kullanarak karakteristik x- ve n- halkalarının bo-yutuna baglı sınıflandırma problemi üzerine çalıstık. Bunun yanında A = (aij)N×NN ×N boyutuna sahip bir A matrisi için,

ui1,x − uix = e∑a+iju

j1+

∑a−iju

j

i, j = 1, 2, . . . , N

formuna sahip yarı ayrık üssel hiperbolik denklemleri de ele alıyoruz. N = 2 durumuiçin yukarıdaki diferansiyel denklemi Darboux integrallenebilir yapacak aij degerle-rini elde ediyoruz.

Anahtar Kelimeler: Darboux integrallenebilir, Karakteristik Lie Halkasi, HiperbolikDenklemler

vi

To my family

vii

ACKNOWLEDGMENTS

My supervisor Assoc. Prof. Dr. Kostyantyn Zheltukhin helped me to a great extent,encouraged me to the very end and made this thesis possible. I take great pleasure inexpressing my sincere gratitude to Kostyantyn Zheltukhin.

My deepest gratitude further goes to my family for being with me in any situation,their encouragement, endless love and trust.

I have been receiving a scholarship about for five years from "Yurt Içi Doktora BursProgramı" of TÜBITAK. I would like to thank to TÜBITAK.

I would like to thank all my close friends whose names only a special chapter couldencompass.

Finally, with deepest feelings I would like to thank my wife Duygu, without whomthis thesis could not exist.

viii

TABLE OF CONTENTS

ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

ÖZ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi

ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . viii

TABLE OF CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix

CHAPTERS

1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 DARBOUX INTEGRABILITY AND CHARACTERISTIC LIE RINGS 5

2.1 Darboux Integrability and Characteristic Lie Rings for Con-tinuous Case . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Darboux Integrability and Characteristic Lie Rings for Semi-Discrete Case . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 ON DARBOUX INTEGRABLE SEMI-DISCRETE HYPERBOLICSYSTEMS OF EXPONENTIAL TYPE . . . . . . . . . . . . . . . . 15

3.1 Semi-Discrete Hyperbolic Systems of Exponential Type . . . 15

3.2 Characteristic x-ring . . . . . . . . . . . . . . . . . . . . . . 17

3.3 Proof of Main Theorem . . . . . . . . . . . . . . . . . . . . 25

3.3.1 The Case X1 = 0 . . . . . . . . . . . . . . . . . . 25

3.3.2 The Case X2 = 0 . . . . . . . . . . . . . . . . . . 26

ix

3.3.3 The Case X3 = 0 . . . . . . . . . . . . . . . . . . 29

3.3.4 Important Formulas For General Case . . . . . . . 33

3.3.5 The Case X4 = 0 . . . . . . . . . . . . . . . . . . 36

3.3.6 The Case X5 = 0 . . . . . . . . . . . . . . . . . . 37

3.3.7 The Case X6 = 0 . . . . . . . . . . . . . . . . . . 42

3.3.8 The Case Xn = 0 n ≥ 7 . . . . . . . . . . . . . 43

4 DARBOUX INTEGRABLE EQUATIONS WITH SMALL DIMEN-SION OF CHARACTERISTIC RINGS . . . . . . . . . . . . . . . . 45

4.1 Characteristic x-ring . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Characteristic n-ring . . . . . . . . . . . . . . . . . . . . . . 48

4.3 Classification of Semi-Discrete Hyperbolic Type EquationsAdmitting Characteristic x-rings and n-rings of Small Di-mensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.3.1 f is Linear With Respect to tx (ftxtx = 0) . . . . 50

4.3.2 f is Nonlinear With Respect to tx (ftxtx 6= 0) . . 53

4.3.2.1 Case 1 : M = t2x . . . . . . . . . . . . 56

4.3.2.2 Case 2 : M =1

tx + P. . . . . . . . . 56

4.3.2.3 Case 3 : M =√t2x + Ptx +Q . . . . 59

5 CONCLUSIONS AND FUTURE WORKS . . . . . . . . . . . . . . 65

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

CURRICULUM VITAE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

x

CHAPTER 1

INTRODUCTION

Throughout the history mathematicians have been interested in methods to solve par-

tial differential equations. Only a few class of nonlinear equations have known meth-

ods of solutions. This reveals the following questions, is there any criteria to deter-

mine if an equation is solvable and how do we solve this equations? To answer this

question we can study properties of equations solved by a given method and try to

find all equations satisfying particular properties. Thus we call an equation integrable

if it satisfies some specially chosen properties. Since it is convenient to use different

properties to describe different classes of equations there is no universal definition of

integrability in the literature. For more about the integrable equations, see [1], [8], [9]

and [13].

The problem of finding all equations of certain type, is known as a "classification

problem", such classification problems are very important in the area of integrable

systems because we want to know all equations solved by a particular method. Many

mathematicians become interested in classification problem since discovery of the

method of inverse scattering transform by Gardner, Green, Kruskal and Miura in

1967 (see [22]). They used inverse scattering transform to solve Koteveg-de Vries

equation. Later it was shown that it was possible to use this method for other non-

linear equations including nonlinear Shrodinger equation (see [40]) and sine-Gordon

equations (see [27]).

Many approaches were introduced attempting to classify equations solved by inverse

1

scattering transform. In 1977 Ablowitz and Segur established a relation between

integrable partial differential equations which are solved by using inverse scattering

transform and the equations that satisfy Painleve test (see [2] and also [6], [39], [29],

[21] and [30]). In 1979 Mikhailov, Sokolov and Shabat introduced the notion of

generalized symmetry which is very useful to classify evolution equations

ut = f(x, u, ux, . . . , unx). (1.1)

It was observed that equations solved by inverse scattering transform have infinite

hierarchy of generalized symmetries. For many classes of the evolution equations

complete classifications were obtained by using symmetry approach (see [11], [12],

[24], [33], [34], [37], [38] , [49] and [50]).

Unfortunately symmetry approach does not work well for hyperbolic equations

uxy = f(x, y, u, ux, uy). (1.2)

In 1915, Darboux suggested definition of integrable hyperbolic equation (1.2) based

on the notion of x- and y-integrals (see [7]).

The functions G = G(x, y, u, uy, uyy, . . . , uny) and F = F (x, y, u, ux, uxx, . . . , umx)

are called x-integral and y-integrals of equation (1.2) if DxG = 0 and DyF = 0

for all solutions of equation (1.2), respectively. We call equation (1.2) is Darboux

integrable if it admits a nontrivial x− and y-integrals. Indeed, if such integrals exist

for an equation (1.2), then the equation can be transformed to a couple of ordinary

differential equations

F (x, y, u, ux, uxx, . . . , umx) = α(x),

G(x, y, u, uy, uyy, . . . , uny) = β(y).

The above ordinary differential equations in principle can be solved for u thus pro-

viding solution of equation (1.2) (see [4],[5] and [10]).

In 1981, Shabat and Yamilov suggested a method for determining Darboux integra-

bility of hyperbolic equations (1.2). They introduced the notion of characteristic Lie

ring for hyperbolic equations (see [23] and [32]). Using this approach, they classify

2

equations of the form

uixy = e∑aiju

j

i, j = 1, 2, . . . , N, (1.3)

(see [32]). It was proved that the system (1.3) is Darboux integrable if and only if the

matrix A = (aij)N×N is a Cartan matrix of a semi-simple Lie algebra. This method

was also successfully apllied to classify other classes of hyperbolic equations (see

[20], [25], [35], [36], [51] and [52]).

In 1999, Adler and Startsev defined Darboux integrability for discrete hyperbolic

equations (see [3]) and in 2005, Habibullin extended definition of characteristic Lie

ring to discrete hyperbolic equations (see [15]). Later in 2007, Habibullin and Pekcan

introduced the the similar definitions for semi-discrete hyperbolic equations

t1x = f(t, t1, tx),∂f

∂tx6= 0, (1.4)

where unknown t = t(n, x) is a function of two independent variables; a discrete

variable n and a continuous variable x (see [17]).

To define Darboux integrability for equation (1.4), which is the main subject of our

study, we introduce the shift operator D, setting Djt(n, x) = t(n + j, x) = tj(n, x)

and derivative operator Dx, setting Djxt(n, x) =

∂j

∂xjt(n, x) = t[j](n, x).

The functions F = F (x, n, t, t1, t−1, t2, t−2, . . . ) and I = I(x, n, t, tx, txx, . . . ) are

x-integral and n-integrals of the chain (1.4) if DxF = 0 and DI = I for all solutions

of (1.4). For more details on the Darboux integrability of equations (1.4), see [14],

[16], [18] and [45].

As we mentioned above there are many classification results for continuous hyper-

bolic equations. But there is no comparable result for discrete and semi-discrete equa-

tions. A restricted classification was done for equations of the form t1,x = tx+d(t, t1)

(see [46]). Some results were obtained toward classification of semi-discrete hyper-

bolic equation with x− and n-ring of small dimension (see [17], [28], [43] and [44]).

For instance it was proven that the only chain with 3 dimensional x-ring and 2 dimen-

sional n-ring is t1x = tx + t1 − t (see [28]).

In 2011, Habibullin, Zheltukhin and Yangubaeva considered discretized form of the

3

equations (1.3) which are called as semi-discrete hyperbolic equations of exponential

type (see [41]). They proposed following discretization that supposedly preserves

integrability

ui1,x − uix = e∑a+iju

j1+

∑a−iju

j

i, j = 1, 2, . . . , N, (1.5)

where the matrix A = (aij)N×N is decomposed into a sum of two triangular matrices

A = A+ + A− with A+ = {a+ij} being upper triangular and A− = {a−ij} being lower

triangular matrices such that all diagonal entries of A+ are equal unity. It was proven

for N = 2 that if A is a Cartan matrix, then the system (1.5) is Darboux integrable.

Also it is hypothesized that the system (1.5) is Darboux integrable if and only if A is

the Cartan matrix of a semi-simple Lie algebra.

We note that discretization hyperbolic equations have many applications in physics

including Toda field equations on discrete space-time, Laplace sequence in discrete

geometry, Q-system, Stokes phenomena in 1D Schrodinger problem and so on. For

more on this subject, see a rewiev paper [26].

This thesis is organized as follows. In Chapter 2, we give the necessary definitions for

Darboux integrability and characteristic Lie rings. In Chapter 3, we prove that when

N = 2 and if the system (1.5) is Darboux integrable then A must be a Cartan matrix

(see also [42]). Finally in Chapter 4, we make a classification of equations (1.4) which

has four dimensional characteristic x-ring and two dimensional characteristic n-ring

(see also[48]).

4

CHAPTER 2

DARBOUX INTEGRABILITY AND CHARACTERISTIC LIE

RINGS

2.1 Darboux Integrability and Characteristic Lie Rings for Continuous Case

The partial differential equations of the form

uxy = f(x, y, u, ux, uy). (2.1)

belong to the class of hyperbolic type differential equations. We introduce notations

uix =∂i

∂xiand uiy =

∂i

∂yi. We define Darboux integrability of equation (2.1) in terms

of x- and y-integrals.

Definition 1 [7] Let n,m <∞.

• A function G = G(x, y, u, uy, uyy, . . . , uny) depending for x, y, u and derivatives of

u with respect to y is x-integral of (2.1) if DxG = 0 on all solutions of (2.1).

• A function F = F (x, y, u, ux, uxx, . . . , umx) depending for x, y, u and derivatives

of u with respect to x is y-integral of (2.1) if DyF = 0 on all solutions of (2.1).

•We call (2.1) Darboux integrable if it admits a nontrivial x-integral and a nontrivial

y-integral.

Here Dx and Dy are operators of differentiation with respect to x and y. In above

equalities the variables u, ux, uy, . . . are assumed to be independent. An x-integral

cannot depend on ux, uxx, . . . . Indeed, suppose G depends on x-derivatives and ukx

is the highest derivative that is G = G(x, y, u, uy, uyy, . . . , uny, ux, . . . , ukx). Then,

DxG = Gx +Gy +Guux + · · ·+Guxuxx + · · ·+Gukxu(k+1)x = 0.

5

The variables u, ux, . . . , u(k+1)x are independent. Note that above expression does not

contain the variable u(k+1)x except the last term. Hence Gukx should be zero, that is

G does not depend on ukx.

From the above definitons it follows that if (2.1) is Darboux integrable, then it can be

transformed to a couple of ordinary differential equations

F (x, y, u, ux, uxx, . . . , umx) = α(x),

G(x, y, u, uy, uyy, . . . , uny) = β(y).

Note that DxG depends on ux.

Example 2 [19] Liouville equation uxy = eu is Darboux integrable and the corre-

sponding x-integral and y-integral are given as,

G = uyy −1

2u2y, F = uxx −

1

2u2x.

Example 3 [19] The equation uxy = eu√u2y − 4 is Darboux integrable and the cor-

responding x-integral and y-integral are given as,

G =uyy − u2y + 4√

u2y − 4, F = uxx −

1

2u2x −

1

2e2u.

It is convenient to define Darboux integrability in terms of characteristic ring. Using

chain rule and the equation (2.1) we can write,

DxG = (∂

∂x+ ux

∂

∂u+ f

∂

∂uy+Dy(f)

∂

∂uyy+ . . . )G = 0

We define a vector field,

X1 =∂

∂x+ ux

∂

∂u+ f

∂

∂uy+Dy(f)

∂

∂uyy+ . . . ,

so that X1G = 0. We also have

∂

∂ux(G) = 0.

sinceG cannot depend on x-derivatives of u. We defineX2 =∂∂ux

. Then we have two

conditions X1G = 0 and X2G = 0. Note that G will be in the kernel of commutators

of the vector fields X1 and X2. Therefore, any vector field in the Lie ring generated

by X1 and X2 annulates G.

6

Definition 4 [32] A Lie ring generated by the vector fields X1 and X2 is called char-

acteristic Lie ring Lx of the equation (2.1) in the direction of x. Characteristic Lie

ring of the equation (2.1) in the direction of y denoted by Ly is defined similarly.

Theorem 5 [23] The equation (2.1) admits x- and y-integrals if and only if charac-

teristic x- and y-rings are finite dimensional, respectively.

Example 6 [13] Liouville equation, uxy = eu has 3 dimensional characteristic Lie

ring Lx with the operators X0, X1 and X2 with a multiplication table,

Lx X1 X2 X3

X1 0 −X3 uxX3 −X1

X2 X3 0 0

X3 X1 − uxX3 0 0

where X1 = Dx, X2 =∂

∂uxand X3 = [X2, X1] =

∂

∂u.

The notion of Darboux integrability is easily extended to the systems of hyperbolic

type,

uixy = f i(x, y, u, ux, uy), i = 1, 2, . . . , n. (2.2)

We introduce notations uijx =∂jui

∂xjand uijy =

∂jui

∂yj. Similar to Definition 1, defini-

tion of Darboux integrability for such systems is given as follows.

Definition 7 Let r, s <∞• A functionG = G(x, y, u1, . . . , un, u1y, . . . , u

ny , . . . , u

1ry, . . . , u

nry) depending on deriva-

tives of ui with respect to y is x-integral of (2.2) ifDxG = 0 for all solutions of (2.2).

• A functionF = F (x, y, u1, . . . , un, u1x, . . . , unx, . . . , u

1sx, . . . , u

nsx) depending on deriva-

tives of ui with respect to x is y-integral of (2.2) if DyF = 0 for all solutions of (2.2).

• We call the system (2.2) is Darboux integrable if it admits n functionally indepen-

dent nontrivial x-integrals and n functionally independent nontrivial y-integrals.

7

From above definitions it follows that if the system (2.2) is integrable then it can be

transformed to a system of ordinary differential equations,

F i(x, y, u1, . . . , un, u1x, . . . , unx, . . . , u

1sx, . . . , u

nsx) = αi(x), i = 1, 2, . . . , n (2.3)

Gj(x, y, u1, . . . , un, u1y, . . . , uny , . . . , u

1ry, . . . , u

nry) = βj(y), j = 1, 2, . . . , n (2.4)

where αi, βj are arbitrary functions.

Example 8 [41] The following system,

uxy = e2u−v

vxy = e−u+2v

is Darboux integrable with x-integrals

G1 = uyy + vyy − u2y + uyvxy − v2yG2 = vyyy + uy(vyy − 2uyy) + u2yvy − uyv2y

and y-integrals

F1 = uxx + vxx − u2x + uxvx − v2xF2 = vxxx + ux(vxx − 2uxx) + u2xvx − uxv2x.

The characteristic rings for a system are defined in the same way as for equations. By

using chain rule we can write

DxG =

(∂

∂x+ u1x

∂

∂u1+ · · ·+ unx

∂

∂un+ f 1 ∂

∂u1y+ . . .

+fn∂

∂uny+Dy(f

1)∂

∂u1yy+ . . .

)G = 0.

We define a vector field,

X1 =∂

∂x+ u1x

∂

∂u1+ · · ·+ unx

∂

∂un+ f 1 ∂

∂u1y+ . . .

+fn∂

∂uny+Dy(f

1)∂

∂u1yy+ . . . .

An x-integral for the system (2.2) does not depend on x derivatives of ui, i = 1, . . . , n.

Hence we define vector fields X2 = ∂∂u1x

, . . . , Xn+1 = ∂∂unx

. Then we have (n + 1)

conditions, that is XiG = 0, i = 1, . . . , (n + 1). Note that G will be in the kernel

of the commutators of the vector fields Xi where i = 1, . . . , (n+ 1). Therefore, any

vector field in the Lie ring composed by Xi’s for i = 1, . . . , (n+ 1) annulates G.

8

Definition 9 The Lie ring defined above is called characteristic Lie ring Lx of the

equation (2.2) in the direction of x. The characteristic Lie ring Ly of the equation

(2.2) in the direction of y can be defined similarly.

Theorem 10 (see [32]) A system of equations (2.2) admits a non-trivial x-integral

and a non-trivial y-integral if and only if its characteristic x-ring Lx and character-

istic y-ring Ly are of finite dimension, respectively.

It can be concluded that the system (2.2) is Darboux integrable if and only if charac-

teristic rings Lx and Ly are finite dimensional.

2.2 Darboux Integrability and Characteristic Lie Rings for Semi-Discrete Case

We consider semi-discrete chains of the given form,

t1x = f(t, t1, tx),∂f

∂tx6= 0 (2.5)

where unknown t = t(n, x) is a function of two independent variables; a discrete

variable n and a continuous variable x [3]. Recall that we defined tk = Dkt and t[k] =

Dkxt. The variables (tk)∞k=−∞ and (t[k])

∞k=1 are considered as independent variables.

Darboux integrability for semi-discrete hyperbolic type equations defined as follows.

Definition 11 [3]

• A function F = F (x, n, t, t1, t−1, t2, t−2, . . . ) depending on shifts of t is x-integral

of the chain (2.5) if DxF = 0 for all solutions of (2.5).

• A function I = I(x, n, t, tx, txx, . . . ) depending on x-derivatives of t is n-integral

of the chain (2.5) if DI = I for all solutions of (2.5) where,

DI = I(x, n+ 1, t1, f, fx, fxx, . . . ) = I(x, n, t, tx, txx, . . . ). (2.6)

In other words, the function I is in the kernel of the operator (D − I)I = 0.

• We call the chain (2.5) Darboux integrable if it admits a nontrivial x-integral

F (x, n, t, t1, t−1, t2, t−2, . . . ) and a nontrivial n-integral I(x, n, t, tx, txx, . . . ).

Remark 12

1. In the same way as in continuous case we can show that an x-integral cannot

9

depend on x-derivatives of t and an n-integral cannot depend on shifts of t.

2. Note that DxI depend on tx and DI depends on shifts of t.

It follows that if the chain (2.5) admits a nontrivial x-integral and a nontrivial n-

integral then by above definitions the following identities,

F (x, n, t, t1, t−1, t2, t−2, . . . ) = α(n)

I(x, n, t, tx, txx, . . . ) = β(x)

must be satisfied for arbitrary α(n) and β(x). This means that the chain (2.5) can be

reduced to a pair of equations; one differential equation and one difference equation.

Example 13 [3] As an example, we consider coupled Riccati equation t1x = tx +

t21 − t2. It is Darboux integrable and the corresponding n-integral and x-integral are

given as,

I = tx − t2, F =(t− t2)(t1 − t3)(t− t3)(t1 − t2)

.

Example 14 [3] The chain t1x =t1 + t

txis Darboux integrable and the correspond-

ing n-integral and x-integral are given as,

I =(txx − 1)2

t2x, F =

(t3 − t1)(t2 − t)(t2 + t1)

.

By shifting (2.5) backward it is possible to write as

t−1x = g(t, t−1, tx) (2.7)

for some appropriate function g. This can be done owing the condition ∂f∂tx6= 0 we

supposed for (2.5).

The characteristic x-ring for the systems is defined in the same way as in continuous

case. We consider vector fieldsD0 = Dx andD1 =∂∂tx

. Then we have two conditions

D0F = 0 and D1F = 0. Note that F will be in the kernel of commutators of the

vector fields D0 and D1. Therefore, any vector field in the Lie ring composed by D0

and D1 annulates F .

Definition 15 Lie ring generated by the vector fields D0 and D1 is called character-

istic Lie ring Lx of the equation (2.5) in the direction of x.

10

Theorem 16 (See [32]) Semi discrete chain (2.5) admits a non-trivial x-integral if

and only if its characteristic x-ring Lx is of finite dimension.

Example 17 [47] For the chain,

t1x = tx +√e2t1 + et1+t + e2t,

characteristic Lie ring Lx is 3 dimensional with the operators D0, D1 and D2 with a

multiplication table,

Lx D0 D1 D2

D0 0 −D2 txD2 −D0

D1 D2 0 0

D2 D0 − txD2 0 0

where D0 = Dx, D1 =∂

∂txand D2 = [D1, D0] =

∞∑k=−∞

∂

∂tk.

The notion of Darboux integrability is easily extended to systems of semi-discrete

type,

ti1x = f i(t1, . . . , tm, t11, . . . , tm1 , t

1x, . . . , t

mx ),

∂f i

∂tx6= 0 i = 1, . . . ,m (2.8)

where unknowns ti = ti(n, x) are functions of two independent variables; a discrete

variable n and a continuous variable x. We defineDjti(n, x) = tij(n, x) = ti(n+j, x)

and Djxt(n, x) = ti[j](n, x) = ∂j

∂xjti(n, x). The variables (tik)

∞k=−∞ and (ti[k])

∞k=1 are

considered as independent variables.

Definition 18

• A function F = F (x, n, t1, . . . , tm, t11, . . . , tm1 , t

1−1, . . . , t

m−1, t

12, . . . ) depending on

shifts of ti’s is x-integral of the chain (2.8) if DxF = 0 for all solutions of (2.8)

• A function I = I(x, n, t1, . . . , tm, t1x, . . . , tmx , t

1xx, . . . ) depending on x-derivatives

of ti(n, x) is n-integral of the chain (2.8) if DI = I for all solutions of (2.8), where

DI = I(x, n+ 1, t11, . . . , tm1 , f

1, . . . , fm, f 1x , . . . , f

nx , f

1xx . . . ).

In other words, the function I is in the kernel of the difference operator (D− I)I = 0.

•We call the chain (2.8) is Darboux integrable if it admitsm functionally independent

nontrivial x-integrals and m functionally independent nontrivial n-integrals.

11

Note that if the chain (2.8) is Darboux integrable, then the following identities,

F i(x, n, t1, . . . , tm, t11, . . . , tm1 , t

1−1, . . . , t

m−1, t

12, . . . ) = αi(n) i = 1, ...,m

Ij(x, n, t1, . . . , tm, t1x, . . . , tmx , t

1xx, . . . ) = βj(x) j = 1, ...,m

should be satisfied for arbitrary functions αi(n) and βj(x).

Example 19 [41] For the system

u1x − ux = eu+u1−v1

v1x − vx = e−u+v+v1

x-integrals are

F1 = e−v+v1 + e−u+u1+v1−v2 + eu1−u2 ,

F2 = e−u+u1 + eu1−u2−v1+v2 + ev2−v3 ,

and n-integrals are given as

I1 = uxx + vxx − u2x + uxvx − v2x,

I2 = u3x + ux(vxx − 2uxx) + u2xvx − uxv2x.

In the same way as in the continuous case, we define vector fields,

D0 = Dx

Di =∂

∂tix, i = 1, . . . ,m.

Then any vector field in the Lie ring generated by Di’s annulates F .

Definition 20 Lie ring generated by the vector fields Di’s i = 0, 1, . . . ,m is called

characteristic Lie ring Lx of the equation (2.8) in the direction of x.

Example 21 [41] For the equation,

u1x − ux = eu+u1−v1

v1x − vx = e−u+v+v1

characteristic Lie-ring Lx consist of the operators ∂∂x

, Y1, Y2, A, B, P1, P2 and T1 =

[P1, P2] where,

Y1 =∂

∂ux, Y2 =

∂

∂vx, A =

∞∑j=−∞

∂

∂uj, B =

∞∑j=−∞

∂

∂vj,

12

P1 =∞∑k=1

(k∑i=1

eui−1+ui−vi)∂

∂uk−∞∑k=1

(k∑i=1

eu−i+u−i+1−v−i+1)∂

∂u−k,

P2 =∞∑k=1

(k∑i=1

e−ui−1+vi−1+vi)∂

∂vk−∞∑k=1

(k∑i=1

e−u−i+v−i+v−i+1)∂

∂v−k.

One can also define the characteristic n-ring such that a nontrivial n-integral exists

if and only if the characteristic n-ring is finite dimensional. Since we are not using

characteristic n-ring in our work we will not give its definition here. The definition is

rather technical and can be found in [17].

13

14

CHAPTER 3

ON DARBOUX INTEGRABLE SEMI-DISCRETE

HYPERBOLIC SYSTEMS OF EXPONENTIAL TYPE

3.1 Semi-Discrete Hyperbolic Systems of Exponential Type

Let A = (aij)N×N be an N ×N matrix and consider hyperbolic systems of exponen-

tial type,

rixy = e∑aijr

j

i, j = 1, 2, . . . , N. (3.1)

The Darboux integrability of such systems is described by the following theorem.

Theorem 22 (see [32]) The system (3.1) is Darboux integrable if and only if A =

(aij)N×N is the Cartan matrix of a semi-simple Lie algebra.

Note that a square matrix A = (aij)N×N is a Cartan matrix if it satifies the follow-

ings,(See [31])

1) aij ∈ {−3,−2,−1, 0, 2},2) aii = 2,

3) aij ≤ 0 when i 6= j,

4) aij = 0 =⇒ aji = 0,

5) There exists a diagonal matrix D such that DAD−1 gives a symmetric and positive

definite quadratic form.

Note that Cartan matrices in two dimension has the form,

A =

2 −1−c 2

, where c = 1, 2, 3. (3.2)

15

We want to consider discretization of (3.1) which preserves integrability. Following

[41], we decompose A into a sum of two triangular matrices A = A+ + A− with

A+ = {a+ij} being upper triangular and A− = {a−ij} being lower triangular matrices

such that all diagonal entries of A+ are equal unity. For example when N = 2, this

decomposition is given by,

A =

a11 a12

a21 a22

, A− =

a11 − 1 0

a21 a22 − 1

, A+ =

1 a12

0 1.

Now we consider discretized form of (3.1). Let ri(n, x), i = 1, 2, . . . , N depend on

a continuous variable x and discrete variable n. We consider semi-discrete chains of

the form (See [41])

ri1,x − rix = e∑a+ijr

j1+

∑a−ijr

j

i, j = 1, 2, . . . , N. (3.3)

For example if we put r1 = u, r2 = v, then for Cartan matrix (3.1), the system (3.3)

becomesu1x − ux = eu+u1−v1

v1x − vx = e−cu+v+v1c = 1, 2, 3. (3.4)

For Darboux integrability of system (3.3) we have the following conjecture.

Conjecture 23 (see [41]) System (3.3) is Darboux integrable if and only if A is a

Cartan matrix .

In [41] it was proven that for 2 × 2 Cartan matrix we have Darboux integrable sys-

tem. Also for the corresponding system (3.4), characteristic x-ring, x-integrals and

n-integrals are fully described.

In this chapter, we classify all A = (aij)2×2 matrixes such that the system (3.3) is

Darboux integrable. In the end, we will show (3.3) is Darboux integrable only if A is

the Cartan matrix. That will complete the proof of Conjecture 23 for the case N = 2.

Our proof is based on characteristic x-ring. Considering characteristic x-ring we also

reproduce results of [41].

Theorem 24 The system of hyperbolic equations of exponential type,

u1x − ux = e(a11−1)u+u1+a12v1

v1x − vx = ea21u+(a22−1)v+v1(3.5)

16

is Darboux integrable if and only if the matrix A = (aij)2×2 is the Cartan matrix.

The proof of the main theorem will be given later. In the next section we consider the

characteristic x-ring to prove this theorem.

3.2 Characteristic x-ring

Consider an arbitrary matrix A = (aij)2x2

A =

a11 a12

a21 a22

, a12, a21 6= 0

Setting r1 = u, r2 = v, the system (3.3) takes form,

u1x − ux = e(a11−1)u+u1+a12v1

v1x − vx = ea21u+(a22−1)v+v1 .(3.6)

Conditions a12, a21 6= 0 are needed because if a12 = 0 or a21 = 0, then (3.6) is

reduced to a set of ordinary differential equations. We assume (3.6) is Darboux inte-

grable and prove that this is possible only if A is the Cartan matrix. The proof will be

based on by constructing x-ring of the system. Let us construct x-ring of the system.

Following Definition 15 the characteristic x-ring is generated by vector fields

Z =∂

∂x+ ux

∂

∂u+ vx

∂

∂v+ u1x

∂

∂u1+ v1x

∂

∂v1+ u(−1)x

∂

∂u−1+ v(−1)x

∂

∂v−1+ . . .

=∂

∂x+

∞∑k=−∞

(ukx

∂

∂uk+ vkx

∂

∂vk

)and

Y1 =∂

∂ux, Y2 =

∂

∂vx.

For convenience we define,

Mi = e(a11−1)ui−1+ui+a12vi , M−i = e(a11−1)u−i+u−i+1+a12v−i+1 ,

Ni = e(a22−1)vi−1+vi+a21ui−1 , N−i = e(a22−1)v−i+v−i+1+a21u−i .

Note that, Mi, M−i, Ni and N−i are never zero and we have

ukx = ux +k∑i=1

Mi, u(−k)x = ux −k∑i=1

M−i, (3.7)

17

vkx = vx +k∑i=1

Ni, v(−k)x = vx −k∑i=1

N−i. (3.8)

We prove the identities (3.7) and (3.8) by using induction. Firstly note that if we shift

(3.6) backward we get,

u(−1)x = ux − e(a11−1)u−1+u+a12v = ux −M−1,

v(−1)x = vx − ea21u−1+(a22−1)v−1+v = vx −N−1.(3.9)

We will prove the identities in (3.7). The other identities be can proved similarly. For

k = 1, the identities in (3.7) follows from (3.6) and (3.9). Now we suppose (3.7) are

true for k = n. That is

unx = ux +n∑i=1

Mi, u(−n)x = ux −n∑i=1

M−i. (3.10)

Then we find

u(n+1)x = u1x +

(n+1)∑i=2

Mi = ux +M1 +

(n+1)∑i=2

Mi = ux +

(n+1)∑i=1

Mi

and

u−(n+1),x = u(−1)x −(n+1)∑i=2

M−i = ux +M−1 +

(n+1)∑i=2

M−i = ux −(n+1)∑i=1

M−i.

Hence the identities (3.7) are true for k = n+ 1, this completes the proof. �

Consider the commutators of Z, Y1 and Y2. They have the following form,

A = [Y1, Z] =∞∑

j=−∞

∂

∂uj, B = [Y2, Z] =

∞∑j=−∞

∂

∂vj.

Next we define A∗ = [A,Z] and B∗ = [B,Z]. Then

A∗ = a11

∞∑k=1

(k∑i=1

Mi

)∂

∂uk− a11

∞∑k=1

(k∑i=1

M−i

)∂

∂u−k+

+ a21

∞∑k=1

(k∑i=1

Ni

)∂

∂vk− a21

∞∑k=1

(k∑i=1

N−i

)∂

∂v−k

and

B∗ = a12

∞∑k=1

(k∑i=1

Mi

)∂

∂uk− a12

∞∑k=1

(k∑i=1

M−i

)∂

∂u−k+

+ a22

∞∑k=1

(Ni)∂

∂vk− a22

∞∑k=1

(k∑i=1

N−i

)∂

∂v−k.

18

For convenience instead of A∗ and B∗ we define the vector fields,

P1 =1

a22a11 − a21a12(a22A

∗ − a21B∗), P2 =1

a22a11 − a21a12(−a12A∗ + a11B

∗).

They have the form,

P1 =∞∑k=1

( k∑i=1

Mi

)∂

∂uk−∞∑k=1

( k∑i=1

M−i

)∂

∂u−k,

P2 =∞∑k=1

( k∑i=1

Ni

)∂

∂vk−∞∑k=1

( k∑i=1

N−i

)∂

∂v−k.

Hence the vector fields,

∂

∂x, Y1, Y2, A, B, P1 and P2

belong to the characteristic x-ring of the system (3.6). Here note that the vector field

Z can be written as linear combination of the above vector fields. Moreover, we have

the following commutator table for this vector fields.

[., .] ∂∂x

Y1 Y2 A B P1 P2

∂∂x

0 0 0 0 0 0 0

Y1 0 0 0 0 0 0 0

Y2 0 0 0 0 0 0 0

A 0 0 0 0 0 a11P1 a21P2

B 0 0 0 0 0 a12P1 a22P2

We can conclude from the above table that except the commutators of P1 and P2 the

other terms do not produce any new vector field. Thus the characteristic x-ring is

finite dimensional if and only if the ring generated by P1, P2 is finite dimensional.

Let A be the ring generated by P1, P2. To describe the structure of A we define a

sequence of linear spaces Xn as follows,

X0 = Lin{P1, P2

},

X1 = Lin{[P1, P2]

},

X2 = Lin{[Pt20 , [Pt21 , Pt22 ]]

}, t20 , t21 , t22 ∈ {1, 2},

...

Xn = Lin{[Ptn0

, [Ptn1, [....[Ptnn−1

, Ptnn]]]]}, tn0 , tn1 , ..., tnn ∈ {1, 2},

19

Lemma 25 The intersection of the linear space Xi and Xj is empty that is Xi

⋂Xj =

0 when i 6= j.

Proof: Consider the vector fields Xi ∈ Xi for i = 1, 2, . . . . Then the vector fields

X1, X2, . . . , Xk are

X1 =∞∑

k=−∞

(∞ ∞∑∑i,j=−∞

A1ijMiNj

)∂

∂uk+

∞∑k=−∞

(∞ ∞∑∑i,j=−∞

B1ijMiNj

)∂

∂vk,

X2 =∞∑

k=−∞

(∞ ∞∑∑i,j=−∞

A2ijMr2i N

R2j

)∂

∂uk+

∞∑k=−∞

(∞ ∞∑∑i,j=−∞

B2ijMr2i N

R2j

)∂

∂vk,

......

Xk =∞∑

k=−∞

(∞ ∞∑∑i,j=−∞

AkijMrki N

skj

)∂

∂uk+

∞∑k=−∞

(∞ ∞∑∑i,j=−∞

BkijMrki N

skj

)∂

∂vk,

where Akij’s and Bkij’s are constants and 1 ≤ rk, sk ≤ k, rk + sk = k + 1 for

k = 1, 2, 3, . . . .

Suppose contrary, Xm

⋂Xn 6= 0 for some m 6= n, then there exist a vector field

C ∈ Xm

⋂Xm such that C = cmXm = cnXn and cm, cn 6= 0 for someXm ∈ Xm and

Xn ∈ Xn. Then the coefficients of∂

∂ukand

∂

∂vkmust be same for cmXm and cnXn.

We must have,

∞ ∞∑∑i,j=−∞

cmAmijMrmi N sm

j =∞ ∞∑∑i,j=−∞

cnAnijMrni N

snj ,

∞ ∞∑∑i,j=−∞

cmBmijMrmi N sm

j =∞ ∞∑∑i,j=−∞

cnBnijMrni N

snj ,

where

M rmi N sm

j = erm(ui+a12vi+(a11−1)ui−1)+sm(vj+a21uj−1+(a22−1)vj−1),

M rni N

snj = ern(ui+a12vi+(a11−1)ui−1)+sn(vj+a21uj−1+(a22−1)vj−1),

are polynomials of exponential powers. Since both of rm = rn and sm = sn cannot

hold at the same time, the polynomials of exponential powers are not same. Hence

we must have cm = cn = 0. This means that the vector field C = 0 and this is a

contradiction with our assumption. Hence the intersection of the linear space Xi and

Xj is empty. �

20

Lemma 25 shows that A is a direct sum of Xn’s.

A = X0 ⊕ X1 ⊕ X2 + . . . .

Lemma 26 If dimA <∞ then the linear space Xn = 0 for some n <∞.

Proof: Suppose contrary, let dimA = p < ∞ for some p ∈ N but A =∞∑n=1

Xn.

By Lemma 25, for arbitrary m ∈ Z+, we can find vector fields Xi ∈ Xi for i =

1, 2, . . . ,m. Hence we can find m linearly independent vector fields in A. Then m

must be less than dimA. But since m is arbitrary this is not possible and that is

a contradiction with our assumption. Therefore, we must have Xn = 0 for some

n <∞. �

To prove our claim we will need the condition Xn = 0 for some n. In this case,

all vector fields in Xn must be zero. To ensure this we must have a condition to

prove a vector field is zero. Thus, the following lemma is very useful to show that

characteristic x-ring is of finite dimension.

Lemma 27 (See [28]) Suppose that the vector field

X =∞∑k=1

(αk

∂

∂uk+ α−k

∂

∂u−k

)+∞∑k=1

(βk

∂

∂vk+ β−k

∂

∂v−k

)satisfies the equality DXD−1 = hX , where h is a function depending on shifts and

derivatives of variables u and v then X = 0.

The next lemma shows how the transformation D(, )D−1 acts on the vector fields A,

B, P1 and P2.

Lemma 28 We have the following identities,

DAD−1 = A, DBD−1 = B,

DP1D−1 = P1 −M1A, DP2D

−1 = P2 −N1B.(3.11)

21

Proof: It is easy to check that DAD−1 = A and DBD−1 = B. Let

DP1D−1 =

∞∑k=1

(αk

∂

∂uk+ α−k

∂

∂u−k

)then

αk = DP1D−1(uk) = DP1(uk−1) = D

(k−1∑k=1

Mi

)=

k−1∑k=1

Mi+1. k = 1, 2, . . . .

Hence we have

αk =

( k∑k=1

Mi

)−M1.

In the same way, we can find the equality for α−k, k = 1, 2 . . . . Thus, we have

DP1D−1 = P1 −M1A. The formula for DP2D

−1 follows similarly. �

Let us we introduce the sequences of the commutators;

X1 : T1 = [P1, P2],

X2 : T2 = [P1, T1], R2 = [P2, T1],

X3 : T3 = [P1, T2], K3 = [P2, T2], H3 = [P1, R2], R3 = [P2, R2].

In general we define

Xn : Tn = [P1, Tn−1], Kn = [P2, Kn−1],

Hn = [P1, Hn−1], Rn = [P2, Rn−1] n = 4, 5, . . . ,

We also define

Xn : Wn = [P2,Wn−1], n ≥ 4, 5 . . . and W3 = T3,

Vn = [P2, Vn−1], n ≥ 5, 6 . . . and V4 = T4,

and

Qij = [Ti, Tj] i, j = 1, 2, 3, . . . ,

We have two lemmas which are very useful for our calculations.

22

Lemma 29 For the vector fields in x-ring we have the following identities,

a) A(M1) = a11M1, A(N1) = a21N1.

b) B(M1) = a12M1, B(N1) = a22N1.

c) P1(M1) =M21 , P1(N1) = 0.

d) P2(M1) = a12M1N1, P2(N1) = N21 .

e) T1(M1) = −a12M21N1, Ti(M1) = 0 i ≥ 2, Ti(N1) = 0, i ≥ 1.

f) R2(M1) = −a12(a12 + 1)M21N

21 i ≥ 2 Ri(N1) = 0, i ≥ 1.

g) Ki(M1) = 0, Ki(N1) = 0, i ≥ 3.

Proof: We prove (e) and (f), the others follow similarly. We start with (e),

T1(M1) = [P1, P2](M1) = P1P2M1 − P2P1M1

= P1(a12M1N1)− P2(M21 ) = −a12M2

1N1.

T1(N1) = [P1, P2](N1) = P1P2N1 − P2P1N1

= P1(N21 )− P2(0) = 0.

T2(M1) = [P1, T1](M1) = P1T1M1 − T1P1M1

= P1(−a12M21N1)− T1(M2

1 ) = 0.

when i ≥ 2, using induction we suppose Tk(M1) = 0 and Tk(N1) = 0. Then,

Tk+1(M1) = [P1, Tk](M1) = P1Tk(M1)− TkP1(M1) = Tk(M21 ) = 0.

Tk+1(N1) = [P1, Tk](N1) = P1Tk(N1)− TkP1(N1) = 0.

Hence they are true for n = k + 1. That completes the proof of part (e). Now we

prove (f), We have

R2(M1) = [P2, T1](M1) = P2T1M1 − T1P2M1

= P2(−a12M21N1)− T1(a12M1N1) = −a12(a12 + 1)M2

1N21 .

R2(N1) = [P2, T1](N1) = P2T1N1 − T1P2N1

= −T1(N21 ) = 0.

23

when i ≥ 2, using induction we suppose Rk(N1) = 0. Then,

Rk+1(N1) = [P2, Rk](N1) = P2Rk(N1)−RkP2(N1) = −Rk(N21 ) = 0.

That means Ri(N1) = 0. Hence it is true for i ≥ 2. �

Lemma 30 We have the following identities;

a) [A, Tn] = (na11 + a21)Tn, e) [B,Kn] = (a12 + na22)Kn,

b) [B, Tn] = (na12 + a22)Tn, f) [B,Wn] = (3a12 + (n− 2)a22)Wn,

c) [B,Rn] = (a12 + na22)Rn, g) [B, Vn] = (4a12 + (n− 3)a22)Vn.

d) [A,Rn] = (a11 + na21)Rn,

Proof: We prove (a) and (b), the others follow similarly. We use induction. For

n = 1 we have

[A, T1] = [A, [P1, P2]] = −[P1, [P2, A]]− [P2, [A,P1]]

= −[P1,−a21P2]]− [P2, a11P1] = (a11 + a21)T1.

Hence (a) is true for n = 1. Now we suppose (a) is true for n = k. That is, we

suppose [A, Tk] = (ka11 + a21)Tk. Then

[A, Tk+1] = [A, [P1, Tk]] = −[P1, [Tk, A]]− [Tk, [A,P1]]

= −[P1,−(ka11 + a21)Tk]]− [Tk, a11P1]

= ((k + 1)a11 + a21)Tk+1.

That means (a) is true for all n.

Now we prove(b). We use induction again. For n = 1 we have

[B, T1] = [B, [P1, P2]] = −[P1, [P2, B]]− [P2, [B,P1]]

= −[P1,−a22P2]]− [P2, a12P1] = (a12 + a22)T1.

Hence (b) is true for n = 1. Now we suppose it is true for n = k. That is, we suppose

[B, Tk] = (ka12 + a22)Tk. Then,

[B, Tk+1] = [B, [P1, Tk]] = −[P1, [Tk, B]]− [Tk, [B,P1]]

= −[P1,−(ka12 + a22)Tk]]− [Tk, a12P1]

= ((k + 1)a12 + a22)Tk+1.

24

That means (b) is true for all n. �

3.3 Proof of Main Theorem

We suppose system (3.5) is Darboux integrable. Then from definition of Darboux

integrability, system (3.5) must admit two functionally independent nontrivial x-

integrals and two functionally independent nontrivial n-integrals. Firstly, we con-

sider x-integrals. Since system (3.5) has a nontrivial x-integral, Theorem 16 implies

characteristic x-ring of the system is finite dimensional, that is dimA < ∞. Then

Lemma 26 implies the linear space Xn = 0 for some n < ∞. Hence we will study

all the cases such that the linear space Xn = 0 and Xk 6= 0 for k = 1, 2, .., n − 1.

Therefore we will find all cases of the system (3.5) that are Darboux integrable. We

will also find the dimension and bases of characteristic x-ring for each Darboux in-

tegrable system. Firstly we will consider the cases X1, X2 and X3 are equal to zero,

then we will prove some general formulas to generalize our result to Xn for all n ∈ N.

We start with the case X1 = 0.

3.3.1 The Case X1 = 0

The only vector field in X1 is

T1 = [P1, P2].

We will show X1 is never zero because the vector field T1 is never zero.

Lemma 31 We have the following identity for the vector field T1

DT1D−1 = T1 + a12N1P1 − a21M1P2 + a21M1N1B. (3.12)

Proof: Using (3.11) we can write,

DT1D−1 = D[P1, P2]D

−1 = [DP1D−1, DP2D

−1] = [P1 −M1A,P2 −N1B]

= [P1, P2]− [P1, N1B]− [M1A,P2] + [M1A,N1B]

= T1 + a12N1P1 − a21M1P2 + a21M1N1B.

25

�

Proposition 32 For the system (3.5), the linear space X1 = 0 is not possible.

Proof: By Lemma 27, T1 = 0 if and only if DT1D−1 = T1 and using Lemma 31 this

is possible if and only if a12 = a21 = 0, which is a contradiction with our assumption.

Hence linear space X1 = 0 is not possible. �

3.3.2 The Case X2 = 0

All possible vector fields in X2 are given as follows,

T2 = [P1, T1], R2 = [P2, T1].

Lemma 33 We have the following identities for the vector fields T2 and R2 in X2,

DT2D−1 = T2 + α2

T1M1T1 + α2

P2M2

1P2 + α2P1M1N1P1 + α2

BM21N1B (3.13)

where

α2T1

= −(2a21 + a11), α2P1

= −a12(a11 + 2a21),

α2P2

= a21(a11 + a21 − 1), α2B = a21(1− a11 − a21),

and

DR2D−1 = R2 + γ2T1N1T1 + γ2P2

N21P2, (3.14)

where

γ2T1 = −(2a12 + a22), γ2P2= a12(1− a22 − a12).

26

Proof: We can write

DT2D−1 = D[P1, T1]D

−1 = [DP1D−1, DT1D

−1]

= [P1 −M1A, T1 + a12N1P1 − a21M1P2 + a21M1N1B]

= [P1, T1] + a12[P1, N1P1]− a21[P1,M1P2] + a21[P1,M1N1B]

−[M1A, T1]− a12[M1A,N1P1] + a21[M1A,M1P2]− a21[M1A,M1N1B].

Using the identities in Lemma 29 and Lemma 30 we get the desired result for T2. We

also have

DR2D−1 = D[P2, T1]D

−1 = [DP1D−1, DT1D

−1]

= [P2 −N1B, T1 + a12N1P1 − a21M1P2 + a21M1N1B]

= [P2, T1] + a12[P2, N1P1]− a21[P2,M1P2] + a21[P2,M1N1B]

−[N1B, T1]− a12[N1B,N1P1] + a21[N1B,M1P2]− a21[N1B,M1N1B].

In the same way, using the identities in Lemma 29 and Lemma 30 we get the desired

result for R2. �

Proposition 34 For the system (3.5) the linear space X2 = 0 if and only if aij’s

satisfy,

a11 = 2, a12 = −1

a21 = −1, a22 = 2

in which case A-ring of the system (3.5) is 3 dimensional with bases P1, P2 and T1.

Proof: Since T2, R2 ∈ X2, they must be zero. Firstly, DT2D−1 = T2 must be

satisfied. From the coefficient of T1 and P2 in (3.13) the followings must be satisfied

α2T1

= −(2a21 + a11) = 0,

α2P2

= a21(a11 + a21 − 1) = 0.

They imply

a11 = 2, a21 = −1. (3.15)

27

Under conditions (3.15), DT2D−1 = T2 is satisfied and Lemma 27 implies T2 = 0.

In addition to that we must also have DR2D−1 = R2, from the coefficient of T1 and

P2 on the right hand side of (3.14) we have

γ2T1 = −(2a12 + a22) = 0,

γ2P2= a21(1− a22 − a12) = 1.

which imply

a12 = −1, a22 = 2 (3.16)

Under the conditions (3.16), DR2D−1 = R2 is satisfied and Lemma 27 implies R2 =

0. This means that all vector fields in X2 are zero. Hence X2 = 0.

Under conditions (3.15) and (3.16) the nonzero vectors fields in X0 and X1 are P1, P2

and T1. Hence, the ring A of (3.5) is 3-dimensional with bases P1, P2 and T1 and we

have the following 8 dimensional characteristic x-ring.

[., .] ∂∂x

Y1 Y2 A B P1 P2 T1

∂∂x

0 0 0 0 0 0 0 0

Y1 0 0 0 0 0 0 0 0

Y2 0 0 0 0 0 0 0 0

A 0 0 0 0 0 2P1 −P2 T1

B 0 0 0 0 0 −P1 2P2 T1

P1 0 0 0 −2P1 P1 0 T1 0

P2 0 0 0 P2 −2P2 −T1 0 0

T1 0 0 0 −T1 −T1 0 0 0

�

Note that we obtained

A =

2 −1−1 2

, (3.17)

which is a Cartan matrix. In this case the system (3.5) reduces to

u1x − ux = eu+u1−v1 ,

v1x − vx = e−u+v+v1 .(3.18)

28

For this system our results reproduce the results of [41]. Also corresponding x-

integrals are [41],

F1 = e−v+v1 + e−u+u1+v1−v2 + eu1−u2 ,

F2 = e−u+u1 + eu1−u2−v1+v2 + ev2−v3 ,

and n-integrals are given as follows

I1 = uxx + vxx − u2x + uxvx − v2x,

I2 = u3x + ux(vxx − 2uxx) + u2xvx − uxv2x.

3.3.3 The Case X3 = 0

All possible vector fields in X3 are given as follows

[P1, T2] = T3, [P2, T2] = K3, (3.19)

[P1, R2] = H3, [P2, R2] = R3. (3.20)

Note that the relation H3 = K3 is satisfied by the Jacobi identity.

Lemma 35 We have the following identities for the vector fields T3 and R3 in X3

DT3D−1 = T3 + α3

T2M1T2 + α3

T1M2

1T1 + α3P1M2

1N1P1

+α3P2M3

1P2 + α3BM

31N1B, (3.21)

where

α3T2

= −3(a11 + a21),

α3T1

= a11(2a11 + 6a21 − 1) + a21(3a21 − 3),

α3P1

= a21a12(6a11 + 3a21 − 3) + a11a12(2a11 − 1),

α3P2

= −a21(1− a11 − a21)(2− 2a11 − a21),

α3B = a21(1− a11 − a21)(2− 2a11 − a21),

and

DR3D−1 = R3 + γ3R2

N1R2 + γ3T1N21T1 + γ3P2

N31P2 + γ3BN

41B, (3.22)

29

where

γ3R2= −(3a12 + 3a22),

γ3T1 = (a12 + 2a22 − 1)(2a12 + a22),

γ3P2= a12(2− 3a22)(1− a12 − a22),

γ3B = a12(1− a12 − a22).

Proof: We can write

DT3D−1 = D[P1, T2]D

−1 = [DP1D−1, DT2D

−1]

= [P1 −M1A, T2 + α2T1M1T1 + α2

P2M2

1P2 + α2P2M1N1P1 + α2

BM21N1B]

= [P1, T2] + α2T1[P1,M1T1] + α2

P2[P1,M

21P2] + α2

P2[P1,M1N1P1]

+α2B[P1,M

21N1B]− [M1A, T2]− α2

T1[M1A,M1T1]

−α2P2[M1A,M

21P2]− α2

P2[M1A,M1N1P1]− α2

B[M1A,M21N1B].

Using the identities in Lemma 29 and Lemma 30 we get the desired result for T3. We

also have

DR3D−1 = D[P2, R2]D

−1 = [DP2D−1, DR2D

−1]

= [P2 −N1B,R2 − (2a12 + a22)N1T1 + a12(1− a22 − a12)N21P2]

= [P2, R2]− (2a12 + a22)[P2, N1T1] + a12(1− a22 − a12)[P2, N21P2]

−[N1B,R2] + (2a12 + a22)[P2, N1T1]− a12(1− a22 − a12)[N1B,N21P2].

In the same way, using the identities in Lemma 29 and Lemma 30 we get the desired

result for R3. �

Lemma 36 If the vector field R3 satisfies R3 = 0, then R2 = 0

Proof: We suppose R2 6= 0 and DR3D−1 = R3 must be satisfied. Then from the

coefficients of R2, T1 and B on the right hand side of (3.22) the followings must be

satisfied,

γ3R2= −(3a12 + 3a22) = 0, (3.23)

γ3T1 = (a12 + 2a22 − 1)(2a12 + a22) = 0, (3.24)

γ3B = a12(1− a12 − a22) = 0. (3.25)

30

Equalities (3.23), (3.24) and (3.25) are satisfied if and only if a12 = 0 and a22 = 0

which is not possible. Hence we find R3 = 0 if and only if R2 = 0. �

Proposition 37 For the system (3.5), the linear space X3 = 0 if and only if aij’s

satisfy,

a11 = 2, a12 = −1,

a21 = −2, a22 = 2.

in which case A-ring of the system (3.5) is 4 dimensional with bases P1, P2, T1 and

T2.

Proof: Assume that X3 = 0 . Since T3, R3 ∈ X3 they must be zero. Since T3 =

[P1, T2] assuming T2 6= 0, DT3D−1 = T3 must be satisfied. From the coefficient of

T2 and T1 on the right hand side of (3.21) we have

α3T2

= −3(a11 + a21) = 0, (3.26)

α3T1

= a11(2a11 + 6a21 − 1) + a21(3a21 − 3) = 0. (3.27)

Solving (3.26) and (3.27) we find

a11 = 2, a21 = −2. (3.28)

Under the conditions (3.28), DT3D−1 = T3 is satisfied and Lemma 27 implies

T3 = 0.

Also, using Lemma 36, R3 = 0 implies R2 = 0, Then from the proof of Proposi-

tion 34, we find

a12 = −1, a22 = 2. (3.29)

Under the conditions (3.28) and (3.29), since R2 = 0 and H3 = K3 the vector fields

in (3.19) and (3.20) are zero. Then we get X3 = 0 and the nonzero vector fields in

Xi, for i = 0, 1, 2 are P1, P2, T1 and T2. Therefore, A ring of the system (3.5) is

4-dimensional with bases P1, P2, T1 and T2 and we have the following 9 dimensional

characteristic x-ring.

31

[., .] ∂∂x

Y1 Y2 A B P1 P2 T1 T2

∂∂x

0 0 0 0 0 0 0 0 0

Y1 0 0 0 0 0 0 0 0 0

Y2 0 0 0 0 0 0 0 0 0

A 0 0 0 0 0 2P1 −2P2 0 2T2

B 0 0 0 0 0 −P1 2P2 T1 0

P1 0 0 0 −2P1 P1 0 T1 T2 0

P2 0 0 0 2P2 −2P2 −T1 0 0 0

T1 0 0 0 0 −T1 −T2 0 0 0

T2 0 0 0 −2T2 0 0 0 0 0

�

Note that we obtained

A =

2 −1−2 2

, (3.30)

which is a Cartan matrix. In this case the system (3.5) reduces to

u1x − ux = eu+u1−v1 ,

v1x − vx = e−2u+v+v1 .(3.31)

For this system our results reproduce the results of [41]. Also corresponding x-

integrals are [41],

F1 = e−u+u1 + e−u1+u2+v2−v3 + eu1−u2−v1+v2 + ev2−v3 ,

F2 = e−v+v1 + e−2u+2u1+v1−v2 + 2e−u+2u1−u2 + e2u1−2u2−v1+v2 + ev2−v3 ,

and n-integrals are given as

I1 = 2uxx + vxx − 2u2x + 2uxvx − v2xx,

I2 = u[4] + ux(v[3] − 2u[3]) + uxx(4uxvx − 2u2x − v2x)

+uxx(vxx − uxx) + vxxux(ux − 2vx) + u4x + u2xv2x − 2u2xvx.

Before considering the next case, we will prove some general formulas for the vector

fields Tn and Rn in Xn.

32

3.3.4 Important Formulas For General Case

Let us find how the transformation D(, )D−1 acts on the introduced sequences of

vector fields.

Lemma 38 We have the following identities for the vector fields Tn and Rn.

DTnD−1 = Tn + αnTn−1

M1Tn−1 + αnTn−2M2

1Tn−2 + . . . , (3.32)

DRnD−1 = Rn + γnRn−1

N1Rn−1 + γnRn−2N2

1Rn−2 + . . . . (3.33)

Proof: We use induction. Equations (3.13) and (3.22) imply the above formulas for

n = 2, 3. Now we suppose they are true for n = k and we check the case n = k + 1,

DTk+1D−1 = [DP1D

−1, DTkD−1]

= [P1 −M1A, Tk + αkTk−1M1Tk−1 + αkTk−2

M21Tk−2 + . . . ]

= Tk+1 + αk+1Tk

M1Tk + αk+1Tk−1

M21Tk−1 + · · ·+ αk+1

TiMk−i+1

1 Ti + . . . ,

where

αk+1Tk

= αkTk−1− (ka11 + a21), (3.34)

αk+1Tk−1

= αkTk−2+ (1− ka11 − a21)αkTk−1

, (3.35)

αk+1Tk−2

= αkTk−3+ (2− ka11 − a21)αkTk−2

, (3.36)

and

DRk+1D−1 = [DP2D

−1, DRkD−1]

= [P2 −N1B,Rk + γkRk−1N1Rk−1 + γkRk−2

N21Rk−2 + . . . ]

= Rk+1 + γk+1Rk

N1Rk + γk+1Rk−1

N21Rk−1 + . . . ,

where

γk+1Rk

= γkRk−1− (ka22 + a12), (3.37)

γk+1Rk−1

= γkRk−2+ (1− ka22 − a12)γkRk−1

, (3.38)

γk+1Rk−2

= γkRk−3+ (2− ka22 − a12)γkRk−2

. (3.39)

33

Hence (3.32) and (3.33) are true for n = k + 1. That means (3.32) is true for n ≥ 2

and (3.33) is true for n ≥ 3. �

Lemma 39 We have the following formulas for the coefficients of the terms ofDTnD−1

and DRnD−1 in (3.32) and (3.33),

αnTn−1= −n(n− 1)

2a11 − na21, n ≥ 2

αnTn−2=

1

24n

(− 8a11 + 12na11 − 4n2a11 − 2a211 + 9na211 − 10n2a211

+ 3n3a211 + 12a21 − 12na21 + 12a11a21 − 24na11a21

+ 12n2a11a21 − 12a221 + 12na221

), n ≥ 3

αnTn−3=

(− 1

48n(n2 − 3n+ 2)(n(n2 − 4n+ 3)a311 + 2a211(n

2(3a21 − 2)

− 7n(a21 − 1) + 2a21 − 3) + 4a11(3na221 − 5na21 + n− 3a221 + 7a21 − 3)

+ 8a21(a221 − 3a21 + 2))

), n ≥ 4

γnRn−1= −n(n− 1)

2a22 − na12, n ≥ 4

γnRn−2=

1

24(−2 + n)

(3n3a222 − 4n2a22(1 + a22 − 3a12) + 12a12(−1 + a22 + a12)

+ n(4a22 + a222 + 12(−1 + a12)a12)

), n ≥ 5

γnRn−3= − 1

48(−3 + n)(−2 + n)

(n4a322 − 2n3a222(2 + a22 − 3a12)

+ 8a12(4− 5a22 + a222 − 6a12 + 3a22a12 + 2a212)

+ n2a22(4 + 6a22 + a222 − 20a12 − 2a22a12 + 12a212) n ≥ 6

+ 2n(a222(−1 + 6a12) + 4a12(2− 3a12 + a212) + a22(−2− 6a12 + 6a212)

)).

34

Proof: Solving the recursive equations (3.34), (3.35), (3.36), (3.37), (3.38) and

(3.39) we get the desired formulas. �

The following two lemmas are very important for our calculations.

Lemma 40 If characteristic x-ring A is finite dimensional then the vector field

R2 = 0 and this is equivalent to a12 = −1 and a22 = 2.

Proof: Lemma 26 implies that the linear space Xn = 0 for some n < ∞. We show

that Xn = 0 will imply the desired results. We use induction, because of Proposi-

tion 32, X1 cannot be zero. So we start from X2 = 0. Since R2 ∈ X2, it should be

zero. Then from the proof of Proposition 34 we get

a12 = −1, a22 = 2. (3.40)

The conditions (3.40) imply DR2D−1 = R2 and Lemma 27 implies R2 = 0.

Suppose now X3 = 0. Since R3 ∈ X3 it must be zero. Then Lemma 36 implies

R2 = 0 and this implies a12 = −1, a22 = 2.

Now suppose it is true for n = k. That is, Xk = 0 implies R2 = 0 and a12 = −1,

a22 = 2. Let n = k + 1 and suppose Xk+1 = 0. Then since Rk+1 ∈ Xk+1, Rk+1 must

be zero, which is equivalent to DRk+1D−1 = Rk+1. Using (3.37) we need γk+1

Rk= 0,

γk+1Rk−1

= 0 and γk+1Rk−2

= 0 and solving them together we find a12 = a22 = 0, which is

a contradiction. Since Rk+1 = [P2, Rk] we must have Rk = 0. Hence by induction

we get the desired result. �

Lemma 41 The vector field Tn = 0 with the condition Tn−1 6= 0 implies the following

conditions on aij’s,

a11 = 2, a21 = 1− n, for n ≥ 2. (3.41)

Proof: When n = 2, the proof of Proposition 34 implies (3.41). When n ≥ 3,

DTnD−1 = Tn must be satisfied. Hence we set the coefficient of Tn−1 and Tn−2

equal to zero in (3.32), that is αnTn−1= 0 and αnTn−2

= 0. Solving them we get the

desired conditions for a11 and a21. �

35

3.3.5 The Case X4 = 0

All possible vector fields in X4 are given as follows,

[Pi, T3] i = 1, 2 , [Pi, R3] i = 1, 2 , [Pi, K3] i = 1, 2 .

Lemma 42 We have the following identity for the vector field Q12 in X4

DQ12D−1 = Q12 + θ12T3N1T3 + θ12K3

M1K3 + θ12T2M1N1T2

+θ12R2M2

1R2 + θ12T1M21N1T1 + . . . , (3.42)

where

θ12T3 = a12, θ12K3= −a21,

θ12T2 = a21(212 + a22), θ12R2= a21(a21 + 1),

θ12T1 = a21(a11 + a21)(a12 + a22).

Proof: We can write

DQ12D−1 = D[T1, T2]D

−1 = [DT1D−1, DT2D

−1]

=[T1 + a12N1P1 − a21M1P2 + a21M1N1B,

T2 + α2T1M1T1 + α2

P2M2

1P2 + α2P1M1N1P1 + α2

BM21N1B

].

Using the identities in Lemma 29 and Lemma 30 we get the desired result for Q12. �

Lemma 43 When X4 = 0, the vector field Q12 = 0 implies T3 = 0.

Proof: Firstly, Lemma 40 implies R2 = 0 and by the case X3 = 0 we have K3 = 0.

Since Q12 = [T1, T2], we suppose T2 6= 0. From the coefficients of T3 on the right

hand side of (3.42) we require

θ12T3 = a12 = 0,

but this is not possible and we find T3 = 0. �

36

Proposition 44 For the system (3.5) the linear space X4 = 0 but Xk 6= 0

for k = 1, 2, 3 is not possible.

Proof: Suppose contrary, X4 = 0 and Xk 6= 0 for k = 1, 2, 3. We know R4, Q12 ∈X4. Then Lemma 40 implies R2 = 0 and Lemma 43 implies T3 = 0. But by Proposi-

tion 37, this two together imply X3 = 0 and this is a contradiction with our assump-

tion. Hence X4 = 0 but Xk 6= 0 for k = 1, 2, 3 is not possible. �

3.3.6 The Case X5 = 0

All possible vector fields in X5 are given as follows,

[Pi, [Pj, T3]] i, j = 1, 2,

[Pi, [Pj, R3]] i, j = 1, 2,

[Pi, [Pj, K3]] i, j = 1, 2.

Lemma 45 We have the following identity for the vector fields Q13 and V5 in X5,

DQ13D−1 = Q13 + θ13T4N1T4 + θ13W4

M1W4 + θ13Q12M1Q12 + θ13T3M1N1T3 + θ13K3

M21K3

+θ13T2M21N1T2 + θ13R2

M31R2 + θ13T1M

31N1T1 + θ13P1

M31N

21P1, (3.43)

where

θ13T4 = a12, θ13W4= −a21,

θ13Q12= −3(a11 + a21), θ13T3 = a21a22 − 3a11a12,

θ13K3= 3a21(a11 + a21), θ13T2 = −3a21(a11 + a21)(a22 + 2a12),

θ13R2= −a21(1 + a21)(−2 + 3a11 + 2a21),

θ13T1 = a21(1 + a21)(−2 + 3a11 + 2a21)(a22 + 2a12),

θ13P1= a21(1 + a21)(−2 + 3a11 + 2a21)a12(−1 + a22 + a12).

and

DV5D−1 = V5 + ϕ5

W4M1W4 + ϕ5

T4N1T4 + ϕ5

T3M1N1T3 + ϕ5

K3M2

1K3

+ϕ5T2M2

1N1T2 + ϕ5R2M3

1R2 + ϕ5T1M3

1N1T1 + . . . , (3.44)

37

where

ϕ5W4

= α4T3, ϕ5

T2= −(2a12 + a22)α

4T2,

ϕ5T4

= −(4a12 + a22), ϕ5R2

= α4T1,

ϕ5T3

= −(3a12 + a22)α4T3, ϕ5

T1= −(a12 + a22)α

4T1− α4

P1,

ϕ5K3

= α4T2.

Proof: We can write

DQ13D−1 = D[T1, T3]D

−1 = [DT1D−1, DT3D

−1]

=[T1 + a12N1P1 − a21M1P2 + a21M1N1B, T3 + α3

T2M1T2

+α3T1M2

1T1 + α3P1M2

1N1P1 + α3P2M3

1P2 + α3BM

31N1B

].

Using the identities in Lemma 29 and Lemma 30 we get the desired result for Q13.

For DV5D−1 we can write

DV5D−1 = D[P2, T4]D

−1 = [DP2D−1, DT4D

−1]

=[P2 −N1B, T4 + α4

T3M1T3

+α4T2M2

1T2 + α4T1M3

1T1 + α4P1M3

1N1P1 + . . .].

In the same way, using the identities in Lemma 29 and Lemma 30 we get the desired

result for V5. �

Lemma 46 When X5 = 0, the vector field V5 = 0 implies,

a) T4 = 0,

b) Q13 = 0.

Proof: a) Firstly, Lemma 40 implies R2 = 0, K3 = 0 and a12 = −1, a22 = 2. Then,

DV5D−1 takes the form,

DV5D−1 = V5 + α4

T3M1W4 − (4a12 + a22)N1T4

−(3a12 + a22)α4T3M1N1T3 + . . . . (3.45)

Since V5 = [P2, T4], we suppose T4 6= 0 and we must have DV5D−1 = V5. Then,

since K3 = 0 the coefficients of T3 on the right hand side of (3.45) must be zero,

38

since 3a12 + a22 6= 0 we get α4T3

= 0. Then, DV5D−1 takes the form,

DV5D−1 = V5 − (4a12 + a22)N1T4 + . . . . (3.46)

Then since the coefficient of T4 is not equal to zero, we get T4 = 0.

b) Using Lemma 40 and (3.43), DQ13D−1 reduces to

DQ13D−1 = Q13 − (2a21 + 3a11)M1Q12 + (2a21 + 3a11)M1N1T3.

By part (a), T4 = 0 and Lemma 41 implies a11 = 2, a21 = −3. Then DQ13D−1 =

Q13 is satisfied and we get the desired result. �

Lemma 47 When X5 = 0, we have the following identities ,

a)W4 = −Q12,

b)W5 = 0,

c) [P1,W4] = 0.

Proof: Since X5 = 0, Lemma 40 implies R2 = 0 and K3 = 0.

a) The vector field [P1, K3] is in X4 and we have

0 = [P1, K3] = [P1, [P2, T2]]] = −[P2, [T2, P1]]− [T2, [P1, P2]]

= [P2, T3] + [T1, T2] = W4 +Q12. (3.47)

Hence we find W4 = −Q12.

b) The vector field W5 = [P2,W4] is in X5 and by part (a)

W5 = [P2,W4] = [P2,−Q12] = −[P2, [T1, T2]] = [T1, [T2, P2]] + [T2, [P2, T1]]

= −[T1, K3] + [T2, R2] = 0.

c) Using Lemma 46 and part (a) we can write

0 = −Q13 = −[T1, T3] = −[T1, [P1, T2]] = [P1, [T2, T1]] + [T2, [T1, P1]]

= −[P1, Q12]− [T2, T2] = −[P1, Q12]

= [P1,W4].

Hence we are done. �

39

Proposition 48 For the system (3.5), the linear space X5 = 0 if and only if aij’s

satisfy,

a11 = 2, a12 = −1,

a21 = −3, a22 = 2,(3.48)

in which case A-ring of the system (3.5) is 6 dimensional with bases P1, P2, T1, T2,

T3 and W4.

Proof: Since R5, V5 ∈ X5, they must be zero. Lemma 40 and Lemma 46 imply

(3.48). It follows that the vector fields [Pi, [Pj, R3]], i, j = 1, 2 and [Pi, [Pj, K3]],

i, j = 1, 2 are all zero. Now we check [Pi, [Pj, T3]], for i, j = 1, 2. Since T4 = 0 we

need to check only the vector fields [P1,W4] and W5 = [P2,W4]. By Lemma 47, they

are also zero. That means all the vector fields in X5 are zero. That is X5 = 0. The

only vector field in X4 which is nonzero is W4. Hence W4 is also a bases vector field

for the characteristic x-ring of system (3.6) in that case.

The nonzero vector fields in Xi, for i = 0, 1, 2, 3 are P1, P2, T1, T2 and T3. Hence

they form a bases for the characteristic x-ring of the system (3.6). Hence, the ring Aof the system (3.6) is 6-dimensional with bases {P1, P2, T1, T2, T3, W4} and we have

the following 11 dimensional characteristic x-ring.

[., .] ∂∂x

Y1 Y2 A B P1 P2 T1 T2 T3 W4

∂∂x

0 0 0 0 0 0 0 0 0 0 0

Y1 0 0 0 0 0 0 0 0 0 0 0

Y2 0 0 0 0 0 0 0 0 0 0 0

A 0 0 0 0 0 2P1 −3P2 −T1 T2 3T3 3W4

B 0 0 0 0 0 −P1 2P2 T1 0 −T3 −W4

P1 0 0 0 −2P1 P1 0 T1 T2 T3 0 0

P2 0 0 0 −2P2 −2P2 −T1 0 0 0 W4 0

T1 0 0 0 T1 −T1 −T2 0 0 W4 0 0

T2 0 0 0 −T2 0 −T3 0 −W4 0 0 0

T3 0 0 0 −3T3 T3 0 −W4 0 0 0 0

W4 0 0 0 −3W4 W4 0 −W4 0 0 0 0

�

40

Note that we obtained

A =

2 −1−3 2

, (3.49)

which is a Cartan matrix. In this case the system (3.5) reduces to,

u1x − ux = eu+u1−v1

v1x − vx = e−3u+v+v1 .(3.50)

For this system our results reproduce the results of [41]. Also corresponding x-

integrals are [41]

F1 = e−u+u1 + e−2u1+2u2+v2−v3 + 2e−u1+2u2−u3 + e−u2+u3+v3−v4

+e2u2−2u3−v2+v3 + eu1−u2−v1+v2 + eu3−u4 ,

F2 = e−v+v1 + 3e−2u+3u1+v1−v2 + 3e−u1+3u2−2u3 + 3e−u+u1+u2−u3 + e−3u+3u1+v1−v2

+3e−u+3u1−2u2−v1+v2 + 3eu1−u3−v1+v2 + e3u1−3u2−2v1+2v2 + e−3u1+3u2+2v2−2v3

+3e−u+u2+v2−v3 + 3e−2u1+3u2−u3+v2−v3 + 2e−v1+2v2−v3 + e3u2−3u3−v2+v3 + ev3−v4 ,

and n-integrals are given as

I1 = uxx +1

3vxx − u2x + uxvx −

1

3v2x,

I2 = u[6] − 2u[5]ux + v[5]ux + 32u[4]u2x − 30u[4]uxvx + 11u[4]v

2x − 40u[4]uxx

+11u[4]vxx + 14v[4]u2x − 15v[4]uxvx +

13

3v[4]v

2x − 10v[4]uxx −

13

3v[4]vxx

+19u2[3] +13

6v2[3] + 16u[3]v[3] − 36u[3]uxxux + 18u[3]uxxvx + 80u[3]vxxux

+− 45u[3]vxxvx − 52v[3]uxxux + 33v[3]uxxvx − 5v[3]vxxux − 64u[3]u3x

+102u[3]u2xvx − 2u[3]uxv

2x + 13u[3]v

3x + 32v[3]u

3x − 58v[3]u

2xvx + 38v[3]uxv

2x

−26

3v[3]v

3x + 66u3xx +

26

3v3xx − 35u2xxvxx − 5uxxv

2xx + 30u2xxu

2x − 18u2xxuxvx

−11

2u2xxv

2x − 34uxxvxxu

2x + 32uxxvxxuxvx − 2uxxvxxv

2x − 2vxxuxvx + 6uxu

4x

−24uxxu3xvx + 25uxxu2xv

2x − 9uxxuxv

3x + uxxv

4x − vxxu4x + 8vxxu

3xvx − 8vxxu

2xv

2x

+2vxxuxv3x − 2u6x + 6u5xvx −

13

2u4xv

2x + 3u3xv

3x −

1

2u2xv

4x.

41

3.3.7 The Case X6 = 0

All possible vector fields in X5 are given as follows,

[Pi, [Pj, [Pk, T3]]] i, j = 1, 2,

[Pi, [Pj, [Pk, R3]]] i, j = 1, 2,

[Pi, [Pj, [Pk, K3]]] i, j = 1, 2.

Lemma 49 We have the following identity for the vector field V6 ∈ X6,

DV6D−1 = V6 − (8a12 + 3a22)N1V5 + α4

T3M1W5

−(4a12 + a22)α5T4N2

1T4 + . . . , (3.51)

Proof: We have

DV6D−1 = D[P2, V5]D

−1 = [DP2D−1, DV5D

−1]

=[P2 −N1B, V5 + ϕ5

W4M1W4 + ϕ5

T4N1T4 + . . .

]= [P2, V5] + ϕ5

W4[P2,W4] + ϕ5

T4[P2, T4]

−[N1B, V5]− ϕ5W4

[N1B,W4]− ϕ5T4[N1B, T4] + . . . ,

Using the identities in Lemma 29 and Lemma 30 we get the desired result for V6. �

Lemma 50 When X6 = 0, the vector field T4 = 0.

Proof: Firstly, note that using Lemma 40 and the proof of Lemma 47 we getW5 = 0.

Since V6 ∈ X6, it must be zero. Using W5 = 0 and Lemma 40, DV6D−1 takes the

form,

DV6D−1 = V6 − (8a12 + 3a22)N1V5 − (4a12 + a22)α

5T4N2

1T4 + . . . . (3.52)

Then by Lemma 27, DV6D−1 = V6 must be satisfied. But since a12 = −1 and

a22 = 2, the coefficient of V5 is not zero so V5 must be zero. Then using Lemma 46

we find T4 = 0. �

42

Proposition 51 For the system (3.5), the linear space X6 = 0 but Xk 6= 0 for k =

1, . . . , 5 is not possible.

Proof: Since R6, V6 ∈ X6, they must be zero. Using Lemma 40 and Lemma 50 we

get R2 = 0 and T4 = 0. But as in proof of Proposition 48 this two together imply

X5 = 0 and this is a contradiction with our assumption. Hence X6 = 0 but Xk 6= 0

for k = 1, . . . , 5 is not possible. �

3.3.8 The Case Xn = 0 n ≥ 7

Lemma 52 The vector field Vn in Xn has the form

DVnD−1 = Vn + ϕnVn−1

N1Vn−1 + ϕnWn−1M1Wn−1, (3.53)

where

ϕnVn−1=

1

2[(42− 8n)a22 + (7n− n2 − 22)a12]. (3.54)

Proof: We use induction. Using (3.51) (3.53) is true for n = 6. Now we suppose it

is true for n = k and we check the case n = k + 1,

DVk+1D−1 = [DP2D

−1, DVkD−1]

=[P2 −N1B, Vk + ϕnVk−1

N1Vk−1 + ϕkWk−1M1Wk−1 + . . .

]= Vk+1 + ϕk+1

VkN1Vk + ϕk+1

WkM1Wk + . . . ,

where

ϕk+1Vk

= ϕkVk−1− (4a12 + (k − 3)a22),

ϕk+1Wk

= ϕkWk−1,

ϕk+1Vk−1

= ϕkVk−2− (4a12 + (k − 3)a22 − 1)ϕkVk−1

,

ϕk+1Wk−1

= ϕkWk−2− (3a12 + (k − 3)a22)ϕ

kWk−1

,

ϕk+1Kk−1

= ϕkKk−2.

Hence (3.53) is true for n = k + 1. That means it is true for all n ≥ 7. �

43

Lemma 53 When n ≥ 7 if Xn = 0, then we have

a) Wi = 0, i ≥ 5,

b) T4 = 0.

Proof: Using Lemma 40, we get R2 = 0 and K3 = 0.

a) Lemma 47 implies W5 = 0. Hence Wi = 0, when i ≥ 5.

b) Since Vn ∈ X6, it must be zero. Using (3.53) with part (a) and (b), DVnD−1 takes

the form

DVnD−1 = Vn + ϕnVn−1

N1Vn−1 + ϕnVn−2N2

1Vn−2 + . . . . (3.55)

Then we use Lemma 27, so DVnD−1 = Vn must be satisfied. We suppose Vn−1 6= 0,

but for a12 = −1 and a22 = 2 the coefficient of Vn−1 on the right hand side of (3.55)

ϕnVn−1=

1

2(n2 − 23n+ 106) (3.56)

is nonzero for n ∈ N. Hence we must have Vn−1 = 0. Continuing in that way we get

V5 = 0 and using Lemma 46 we get T4 = 0. �

Proposition 54 , For the system (3.5), when n ≥ 7 the linear space Xn = 0 but

Xk 6= 0, k = 1, 2, ..., (n− 1) is not possible.

Proof: If Xn = 0, Lemma 40 imply R2 = 0 and from Lemma 53 we get T4 = 0.

But in that case from the proof of Proposition 48, it follows that X5 = 0 and this is a

contradiction with our assumption.

Hence for the system (3.6) when n ≥ 7, Xn = 0 but Xk 6= 0, k = 1, 2, ..., (n − 1) is

not possible. �

As a result, Proposition 32, Proposition 34, Proposition 37, Proposition 44, Proposi-

tion 48, Proposition 51 and Proposition 54 imply that, if characteristic x-ring is finite

dimensional, then A should be a Cartan matrix in the form (3.2). In [41], when A is a

Cartan matrix in the form (3.2) two functionally independent nontrivial n-integrals of

the corresponding system were constructed. Hence we can conclude that if the sys-

tem (3.5) is Darboux integrable, then A must be a Cartan matrix and that completes

the proof of Theorem 24.

44

CHAPTER 4

DARBOUX INTEGRABLE EQUATIONS WITH SMALL

DIMENSION OF CHARACTERISTIC RINGS

In this chapter, we consider some classification problems for the chains of the form

t1x = f(x, t, t1, tx). (4.1)

In particular we consider chains with x- and n-ring of small dimension.

4.1 Characteristic x-ring

For semi-discrete chains of type (4.1), x-ring is generated as described in Chapter 2

by the vector fields,

D0 =∂

∂x+ tx

∂

∂t+ f

∂

∂t1+ g

∂

∂t−1+D(f)

∂

∂t2+D−1(g)

∂

∂t−2+ . . .

D1 =∂

∂tx.

where g = t−1,x. To be able to construct characteristic x-ring, we consider the com-

mutators of D0 and D1. Let us define sequences of vector fields

C1 = [D0, D1] and Cn = [X,Cn−1] n = 2, 3, . . . ,

and

Z1 = [D0, C1] and Zn = [D0, Zn−1] n = 2, 3, . . . .

45

This vector fields have form [28], [43]

C1 =∂

∂t+ ftx

∂

∂t1+ gtx

∂

∂t−1+ (f1)tx

∂

∂t2+ (g−1)tx

∂

∂t−2. . .

C2 = ftxtx∂

∂t1+ gtxtx

∂

∂t−1+ (f1)txtx

∂

∂t2+ (g−1)txtx

∂

∂t−2. . .

Z1 = (ftxx + txftxt + fftxt1 − ft − ftxft1)∂

∂t1+

+(gtxx + txgtxt + ggtxt1 − gt − gtxgt1)∂

∂t−1+ . . . .

In general we have

Cn =∞∑i=1

((D1)

n(fi−1)∂

∂ti+ (D1)

n(g−i+1)∂

∂t−i

),

where f0 := f and g0 := g. In what follows we use some commutator relations

between vector fields.

Lemma 55 (See [28]) The vector field D0 satisfies a commutation relation

DD0 =1

ftxD0D. (4.2)

Assumption that characteristic x-ring has a small dimension puts restrictions on the

function f . Such restrictions were found in [28] and [43].

Lemma 56 (See [28]) For the chain (4.1), dimension of the characteristic x-ring is

three if and only if the function f satisfies

ftxtx = 0 (4.3)

and

− txftxtf 2tx

− fftxt1f 2tx

+ftf 2tx

+ft1ftx

= 0. (4.4)

In this case, characteristic x-ring is generated by the vector fields D0, D1 and C1,

which form the following commutator table

Lx D0 D1 C1

D0 0 C1 0

D1 −C1 0 0

C1 0 0 0

46

When dimension of the characteristic x-ring is 4, we have to consider two different

cases, ftxtx 6= 0 and ftxtx = 0.

Lemma 57 (See [43]) Suppose ftxtx 6= 0 then the chain (4.1) has characteristic x-

ring of dimension four if and only if the following conditions hold

D

(ftxtxtxftxtx

)=ftxtxtxftx − 3f 2

txtx

ftxtxf2tx

, (4.5)

D

(fxtx + txftxt + fftxt1 − ft − ftxft1

ftxtx

)=

fxtx + txftxt + fftxt1 − ft − ftxft1ftxtx

ftx − (fx + txft + ft1 .f).

(4.6)

The characteristic x-ring is generated by the vector fieldsD0, D1, C1, C2 and we have

the following commutator table:

Lx D0 D1 C1 C2

D0 0 C1 C2 µC2

D1 −C1 0 λC2 ρC2

C1 −C2 −λC2 0 ηC2

C2 −µC2 −ρC2 ηC2 0

where ρ = λµ+D0(λ) and η = D0(ρ)−D1(µ).

Lemma 58 (See [43]) Supose ftxtx = 0 then the chain (4.1) has the characteristic

x-ring of dimension four if and only if the following condition hold

D

(K(m)

m−m+

ftftx

)=K(m)

m+m− ft1 , (4.7)

where K is the following vector field

K =∂

∂x+ tx

∂

∂t+ f

∂

∂t1+ . . . ,

and m =−(fxtx + txftxt + fftxt1) + ft + ftxft1

ftx. In this case, characteristic x-ring

is generated by the vector fields D0, D1, C1, Z1 and we have the following commuta-

tor table

47

Lx D0 D1 C1 Z1

D0 0 C1 0 0

D1 −C1 0 Z1 αZ1

C1 0 −Z1 0 D0(α)Z1

Z1 0 αZ1 −D0(α)Z1 0

where Z2 = αZ1.

When characteristic x-ring has dimension 4, one can obtain further restrictions on

f(x, t, t1, tx).

Lemma 59 (See [43]) Let the chain (4.1) have four dimensional characteristic x-

ring. Then f has the following form

f =M(x, t, tx)A(x, t, t1) + txB(x, t, t1) + C(x, t, t1), (4.8)

where M, A, B and C are some functions.

Lemma 60 (See [43]) Let the chain (4.1) have four dimensional characteristic x-

ring and ftxtx 6= 0. Then

Df = −H1(x, t, t1, t2)tx +H2(x, t, t1, t2)f +H3(x, t, t1, t2), (4.9)

where H1, H2 and H3 are some functions.

4.2 Characteristic n-ring

For semi-discrete chains of type (4.1) the characteristic n-ring is generated, as de-

scribed in Chapter 2, by the vector fields,

Y0 =∂

∂t1, Y−0 =

∂

∂t−1,

Yj = D−j∂

∂t1Dj, j = 1, 2, . . . ,

Y−j = Dj ∂

∂t−1D−j, j = 1, 2, . . . ,

Xj =∂

∂t−j, j = 1, 2, . . . .

48

When dimension of characteristic n-ring is 2 the restriction on the function f are

obtained in [28]. We have the following condition.

Lemma 61 (See [28]) The dimension of the characteristic n-ring is 2 if and only if f

satisfied the following condition

D

(ftftx

)= −ft1 . (4.10)

In this case, the characteristic n-ring is generated by the vector fields X1 and Y1 and

this vector fields commute.

4.3 Classification of Semi-Discrete Hyperbolic Type Equations Admitting Char-

acteristic x-rings and n-rings of Small Dimensions

The case of three dimensional characteristic x-ring and two dimensional characteristic

n-ring is considered in [28] and it is found that the chain (4.1) must have the form,

t1x = tx + t1 − t.

In this section we want to classify all chains (4.1) such that characteristic x-ring is four

dimensional and characteristic n-ring is two dimensional. To make this classification,

we use Lemma 57, Lemma 58 and Lemma 61. Firstly we remark that using equality

(4.10) in Lemma 61 we get∂

∂t1

(ftftx

)= 0 (4.11)

since ft1 does not depend on t2. In the same way, using the identities (4.6) and (4.7)

we find∂

∂t1m = 0,

∂

∂t1m = 0. (4.12)

where m =fxtx+txftxt+fftxt1−ft−ftxft1

ftxtx.

Note that if the chain (4.1) has four dimensional characteristic x-ring then by Lemma 59

the function f has form

f = A(x, t, t1)M(x, t, tx) +B(x, t, t1)tx + C(x, t, t1).

For simplicity we assume that the function M depends only on tx and f does not

depend on x explicitly. That is we study equations of the form

t1x = A(t, t1)M(tx) +B(t, t1)tx + C(t, t1). (4.13)

We have to consider two cases: ftxtx = 0 and ftxtx 6= 0.

49

4.3.1 f is Linear With Respect to tx (ftxtx = 0)

Since f is linear in tx we have

f(t, t1, tx) = A(t, t1)tx +B(t, t1) (4.14)

and equation (4.1) becomes

t1x = A(t, t1)tx +B(t, t1). (4.15)

Lemma 62 For the chain (4.15) the dimension of the characteristic n-ring is two if

and only if function f has the form

f(t, t1, tx) =c1γ(t)

γ(t1)tx +

c2γ(t1)

− c1γ(t)σ(t)

γ(t1)+ σ(t1), (4.16)

where γ and σ are some functions of the given variables and c1, c2 are some constants.

Proof: If characteristic n-ring is two dimensional, then f must satisfy the condition

(4.10) in Lemma 61

D

(ftftx

)= −ft1 . (4.17)

Then using (4.11) we get that

ftftx

=AtAtx +

Bt

A. (4.18)

which does not depend on t1. ThusAtAtx and

Bt

Adoes not depend on t1. Hence

AtA

= α(t) for some function α(t) and we find,

A(t, t1) = γ(t)ϕ(t1)

for some functions γ and ϕ. Then,

Bt(t, t1)

A(t, t1)=

Bt(t, t1)

γ(t)ϕ(t1)= α(t)

and we find

B(t, t1) = l(t)ϕ(t1) + σ(t1).

50

Then (4.15) and (4.18) reduce to

f(t, t1, tx) = γ(t)ϕ(t1)tx + l(t)ϕ(t1) + σ(t1)

and

ftftx

=γ′(t)

γ(t)tx +

l′(t)

γ(t). (4.19)

Using (4.19) and (4.17) we obtain

γ′(t1)

γ(t1)(γ(t)ϕ(t1)tx + l(t)ϕ(t1) + σ(t1)) +

l′(t1)

γ(t1)= γ(t)ϕ′(t1)tx + l(t)ϕ′(t1) + σ′(t1).

(4.20)

By comparing the coefficients of tx in (4.20) we get

γ′(t1)γ(t)ϕ(t1)

γ(t1)= −γ(t)ϕ′(t1) ⇐⇒ γ′(t1)

γ(t)+ϕ′(t1)

ϕ(t)= 0.

Then we find,

ϕ(t1) =c1γ(t1)

where c1 is a constant. (4.21)

Using (4.21) in (4.20) we obtain

γ′(t1)

γ(t1)σ(t1) +

l′(t1)

γ(t1)= −σ′(t1) ⇐⇒ γ′(t1)σ(t1) + γ(t1)σ

′(t1) + l′(t1) = 0.

(4.22)

Solving (4.22), we find

l(t) = −γ(t)σ(t) + c2, where c2 is a constant. (4.23)

Substituting ϕ and l into (4.19) we get equation (4.16) and condition (4.10) is satisfied

for this f .

t1x = c1γ(t)

γ(t1)tx − c1

γ(t)

γ(t1)σ(t) + c2

1

γ(t1)+ σ(t1), c1, c2 : constants (4.24)

where γ and σ are some functions of one variable. �

Let us introduce a new variable τ

τ = L(t),

such that L′(t) = γ(t). Then (4.24) becomes

τ1x = c1τx + c2 − c1Q(τ) +Q(τ1) (4.25)

51

where Q is function of one variable. For the above equation, it can be shown that

(4.10) is satisfied. Hence for the chain (4.25) dimension of characteristic n-ring is

still remains two.

For the chain (4.25) characteristic x-ring must be four dimensional. Therefore, we

check the condition (4.7) and we have the following lemma.

Lemma 63 The equation (4.25) has four dimensional characteristic x-ring if and

only if

Q(τ) = A1τ2 + A2τ or Q(τ) = A1e

ατ + A2e−ατ (4.26)

for some constants A1, A2 and α.

Proof: In this case we take the function f = c1τx + c2 − c1Q(τ) + Q(τ1) and we

check (4.7) for f . We get

c1τx + c2 − c1Q(τ) +Q(τ1)

Q′(τ2)−Q′(τ1)(c1Q

′′(τ2)−Q′′(τ1))+

+Q′′(τ2)(c2 − c1Q(τ1) +Q(τ2))

Q′(τ2)−Q′(τ1)−Q′(τ2) +Q′(τ1) =

c1Q′′(τ1)−Q′′(τ)

Q′(τ1)−Q′(τ)τx +

Q′′(τ1)(c2 − c1Q(τ) +Q(τ1))

Q′(τ1)−Q′(τ)+Q′(τ1)−Q′(τ).

By comparing coefficients of τx in the above equality we get

c1D

(c1Q

′′(τ1)−Q′′(τ)Q′(τ1)−Q′(τ)

)=c1Q

′′(τ1)−Q′′(τ)Q′(τ1)−Q′(τ)

.

Let F (t, t1) =c1Q

′′(t1)−Q′′(t)Q′(t1)−Q′(t)

, then (4.27) reduces to,

c1D(F (t, t1)) = F (t, t1). (4.27)

Since right hand side of (4.27) does not depend on t2, left hand side of (4.27) also does

not depend on t2. That means F does not depend on t1. Hence we get, c1D(F (t)) =

F (t). Since F does not depend on t1, D(F (t)) does not depend on t1. Then, F does

not depend on t. Hence c1 = 1 or F = 0. When F = 0, c1Q′′(τ1) − Q′′(τ) = 0.

Again this implies c1 = 1. Therefore, the system (4.25) reduces to,

τ1x = τx + d(τ, τ1). (4.28)

in which case complete list of Darboux integrable chains is given at [46] (also, their

x and n characteristic rings). By checking the list of Darboux integrable equations of

52

this form at [46] we conclude that Q should be in the form of (4.26). �

Returning the variable t and using each form of Q given by (4.26), we have the fol-

lowing theorem.

Theorem 64 Let f be a linear function with respect to tx. The equation (4.15) has

four dimensional characteristic x-ring and two dimensional characteristic n-ring if

and only if

f =γ(t)

γ(t1)tx −

γ(t)

γ(t1)σ(t) + σ(t1),

where functions γ and σ satisfy one of the following relations

(γ(t)σ(t))′ = γ(t)√B1 +B2γ(t)σ(t) (4.29)

or

(γ(t)σ(t))′ = γ(t)√B1 +B2(γ(t)σ(t))2 (4.30)

with B1, B2 being arbitrary constants.

Proof: Definition of τ implies Q(τ) = σ(t)γ(t) and Lemma 63 implies Q has the

following forms

Q(τ) = A1τ2 + A2τ or Q(τ) = A1e

ατ + A2e−ατ .

In the first case, Q′(τ) = 2A1τ + A2 which implies,

(Q′(τ))2 = 4A1(A1τ2 + A2τ) + A2

2 = 4A1Q(τ) + A22.

Since (Q′(τ))2 = ((γ(t)σ(t))′)2 we get (4.29). Equality (4.30) follows similarly. �

4.3.2 f is Nonlinear With Respect to tx (ftxtx 6= 0)

We suppose that f is nonlinear in tx and

f(t, t1, tx) = A(t, t1)M(tx) +B(t, t1)tx + C(t, t1) (4.31)

so equation (4.1) becomes

t1x = A(t, t1)M(tx) +B(t, t1)tx + C(t, t1). (4.32)

Using that characteristic n-ring is two dimensional, we have the following lemma.

53

Lemma 65 Let equation (4.32) has characteristic n-ring of dimension two and (Mtx 6=constant) be a non-linear function, then M satisfies

M ′ = −α2M + α4tx + α6

α1M + α3tx + α5

. (4.33)

where α1M + α3tx + α5 6= 0.

Proof: If characteristic n-ring is two dimensional, then f must satisfy the condition

(4.11). Using (4.31) for f we obtain(AtM + txBt + Ct

AM ′ +B

)t1

=

(Att1M + txBtt1 + Ctt1)(AM′ +B)− (At1M

′ +Bt1)(AtM + txBt + Ct)

(AM ′ +B)2= 0.

Rearranging terms in above expression we obtain

M ′(α1M + α3tx + α5) = −(α2M + α4tx + α6), (4.34)

where the constants αi, i = 1, 2, . . . , 5 are given by,

α1 = Att1A− At1At,

α2 = Att1B −Bt1At,

α3 = Btt1A−BtAt1 ,

α4 = Btt1B −Bt1Bt,

α5 = Ctt1A− At1Ct.

If α1M + α3tx + α5 = 0, then M is linear in tx, which implies we are in the case

ftxtx = 0. Hence we assume α1M + α3tx + α5 6= 0 and we obtain the result,

M ′ = −α2M + α4tx + α6

α1M + α3tx + α5

. (4.35)

�

Using (4.35) we can write M ′ in terms of M . The next lemma enables us to write

DM in terms of M .

Lemma 66 If characteristic x-ring is four dimensional and ftxtx 6= 0, then M satis-

fies

DM = Q1(t, t1, t2)M +Q2(t, t1, t2)tx +Q3(t, t1, t2), (4.36)

for some functions Q1, Q2 and Q3.

54

Proof: Since we assume that the dimension of characteristic x-ring is four and

ftxtx 6= 0, by Lemma 60 we have

Df = −H1(t, t1, t2)tx +H2(t, t1, t2)f +H3(t, t1, t2),

where H1, H2 and H3 are some functions. Thus,

D(AM +Btx + C) = −H1tx +H2(AM +Btx + C). (4.37)

We can write (4.37) in the form,

DM = Q1(t, t1, t2)M +Q2(t, t1, t2)tx +Q3(t, t1, t2). (4.38)

where Q1, Q2 and Q3 are some functions. �

The expressions (4.35) and (4.38) enable us to prove the following lemma.

Lemma 67 Let equation (4.32) have characteristic n-ring of dimension two then ei-

ther M =1

tx + Por M =

√t2x + Ptx +Q or M = t2x.

Proof: Note that by (4.2) we have the identity DD0 =1

ftxD0D, where D0 =

∂

∂tx.

Therefore

DD0M =1

ftxXD0M,

DM ′ =1

AM ′ +B

∂

∂txM,

−D(α2M + α4y + α6

α1M + α3y + α5

)=

1

AM ′ +B(Q1M +Q2tx +Q3). (4.39)

We use (4.35) and (4.38) and obtain

− α2(Q1M +Q2tx +Q3) + α4(AM +Btx + C) + α6

α1(Q1M +Q2tx +Q3) + α3(AM +Btx + C) + α5

=

Q2 −Q1α2M + α4tx + α6

α1M + α3tx + α5

B − Aα2M + α4tx + α6

α1M + α3tx + α5

where Dαk = αk. Then

(α2Q1 + α4A)M + (α2Q2 + α4B)tx + (α2Q3 + α4C + α6)

(α1Q1 + α3A)M + (α1Q2 + α3B)tx + (α1Q3 + α3C + α5)=

(Q1α2 −Q2α1)M + (Q1α4 −Q2α3)tx + (Q1α6 −Q2α5)

(Aα2 −Bα1)M + (Aα4 −Bα3)tx + (Aα6 −Bα5).

55

The above equality can be reduced to

R1M2 − (R2tx +R3)M + (R4t

2x +R5tx +R6) = 0, (4.40)

where Ri, i = 1, 2, . . . , 6 are some functions. In the case R1 = 0, from (4.40) we find

that

M =R4t

2x +R5tx +R6

R2tx +R3

. (4.41)

In the case R1 6= 0 we find

M =(R2tx +R3)±

√(R2tx +R3)2 − 4R1(R4t2x +R5tx +R6)

2R1

. (4.42)

Substituting M given by (4.41) into f = AM + Btx + C we can redefine M,B,C

so that

M = t2x or M =1

tx + P.

For M given by (4.42) we can also redefine M,B,C so that M =√t2x + Ptx +Q.

�

Now we consider each case of M obtained in the above lemma separately.

4.3.2.1 Case 1 : M = t2x

Lemma 68 The equation (4.32) cannot have four dimensional characteristic x-ring

if M = t2x.

Proof: In this case f takes the form

f = A(t, t1)t2x +B(t, t1)tx + C(t, t1).

It can be checked that the condition (4.5) is not satisfied. Hence equation (4.32) can-

not have 4 dimensional characteristic x-ring if M = t2x. �

4.3.2.2 Case 2 : M =1

tx + P

In this case we have the following lemma.

56

Lemma 69 Let M =1

tx + Pand equation (4.32) has four dimensional characteris-

tic x-ring and two dimensional characteristic n-ring. Then chain (4.32) is either

t1x =c∗η(t)η(t1)

txor t1x =

c∗ec∗∗(t+t1)

tx + P− P, (4.43)

where c∗ and c∗∗ are some constants.

Proof: In this case f takes the form

f(t, t1, tx) =A(t, t1)

tx + P+B(t, t1)tx + C(t, t1), (4.44)

with

ftx =B(tx + P )2 − A

(tx + P )2,

ftxtx =2A

(tx + P )3,

ftxtxtx = − 6A

(tx + P )4.

By checking condition (4.5) for f we obtain

3(tx + P )

B(tx + P )2 + (C + P −BP )(tx + P ) + A=

3(tx + P )(B(tx + P )2 + A)

(B(tx + P )2 − A)2.

This equality reduces to

B(C + P −BP )(tx + P )3 + 4AB(tx + P )2 + A(C + P −BP )(tx + P ) = 0.

For this equality by checking the coefficients of (tx + P )k, k =1,2,3 we find

(tx + P )3 : B(C + P −BP ) = 0, (4.45)

(tx + P )2 : 4AB = 0, (4.46)

(tx + P ) : A(C + P −BP ) = 0. (4.47)

Firstly, note that A(t, t1) 6= 0. Otherwise ftxtx = 0. By solving (4.45), (4.46) and

(4.47) together we find that

B = 0,

C = −P.

Thus (4.44) takes the form

f(t, t1, tx) =A(t, t1)

tx + P− P. (4.48)

57

Using the condition (4.10) we obtain

At1(t1, t2)A(t, t1)

A(t1, t2)(tx + P )=At1(t, t1)

tx + P

or

At1(t1, t2)

A(t1, t2)=At1(t, t1)

A(t, t1). (4.49)

Then using (4.49)At1(t1, t2)

A(t1, t2)does not depend on t2, so we have

∂

∂t2

∂

∂t1lnA(t1, t2) = 0. Also since

At1(t, t1)

A(t, t1)does not depend on t we have

∂

∂t

∂

∂t1lnA(t, t1) = 0. Hence we get

A(t, t1) = ϕ(t)η(t1).

Substituting A into (4.49) we obtain

ϕ′(t1)

ϕ(t1)=η′(t1)

η(t1)=⇒ d

dt1

(lnϕ(t1)

η(t1)

)= 0 =⇒ ϕ(t1) = c∗η(t1).

Then we have

f(t, t1, tx) =c∗η(t)η(t1)

tx + P− P, where P is a constant.

Using condition (4.12)

m =

(− Pη′(t)

2η(t)+Pη′(t1)

2η(t1)

)(tx + P )− (tx + P )2η′(t)

2η(t)

and it should not depend on t1. This is possible in two cases

Case 1 : P = 0,

Case 2 : η′(t1) = c∗∗η(t1) =⇒ η(t1) = ec∗∗t1 .

For each of these cases condition (4.12) is satisfied and we have

Case 1 : f(t, t1, tx) =c∗η(t)η(t1)

tx,

Case 2 : f(t, t1, tx) =c∗ec

∗∗(t+t1)

tx + P− P, c∗, c∗∗ : constants.

It can be checked that the chains with such functions f have 2 dimensional character-

istic n-ring and 4 dimensional characteristic x-ring. �

58

4.3.2.3 Case 3 : M =√t2x + Ptx +Q

For this case we have the following lemma.

Lemma 70 Let M =√t2x + Ptx +Q and equation (4.32) has four dimensional

characteristic x-ring and two dimensional characteristic n-ring. Then equation (4.32)

takes form

t1x =√B2 − 1

√t2x + Ptx +Q+Btx +

P

2(B − 1). (4.50)

where B, Q, P are constants and B2 6= 1.

Proof: In this case f takes the form

f = A(t, t1)√t2x + Ptx +Q+B(t, t1)tx + C(t, t1). (4.51)

Applying condition (4.5) to f we find,

−6BP + 12tx + 12A

√t2x + txP +Q(

AP + 2Atx + 2B√t2x + txP +Q

)2 =

−3P − 6C − 6Btx − 6A√t2x + Ptx +Q

2Q+ 2(C +Btx + A

√t2x + Ptx +Q

)(C + P +Btx + A

√t2x + Ptx +Q

) ,or

4(Q+

(C +Btx + A

√t2x + Ptx +Q

)(C + P +Btx + A

√t2x + Ptx +Q

))×

×(BP + 2tx + 2A

√t2x + txP +Q

)=(

AP + 2Atx + 2B√t2x + txP +Q

)2 (P + 2C + 2Btx + 2A

√t2x + Ptx +Q

)Then by comparing coefficients of

(√t2x + Ptx +Q

)i(tx

)j, for i, j = 0, 1, 2, we

find

AB(2C + P −BP ) = 0, (4.52)

A(−4C2 − 4CP + A2(P 2 − 4Q) + 4(B2 − 1)Q) = 0, (4.53)

(A2 +B2)(−2C +BP − P ) = 0, (4.54)

−2A2P (2C + P ) + 4B3Q−

−B(4C2 + 4CP + 4Q+ A2(−3P 2 + 4Q)) = 0, (4.55)

A2(2C + P )(P 2 − 8Q) +

+4B2(2C + P )Q− 4BP (C2 + CP +Q− A2Q) = 0. (4.56)

59

Equalities (4.52) and (4.54) are satisfied if

2C = PB − P. (4.57)

Equality (4.53) is satisfied if

4C2 + 4CP = A2(P 2 − 4Q)− 4Q+ 4B2Q, (4.58)

equalities (4.57) and (4.58) are satisfied in two cases

Case 1 : 2C = PB − P and A2 + 1 = B2,

Case 2 : 2C = PB − P and P 2 = 4Q.

In the first case, M reduces to M = tx +P2

which is linear with respect to tx. There-

fore, we study only the Case 2. Thus we have

f =√B(t, t1)2 − 1

√t2x + Ptx +Q+B(t, t1)tx +

P

2(B(t, t1)− 1), (4.59)

where B 6= ±1. In the same way we check that condition (4.11) in the form

(Dftftx

)2

− (ft1)2 = 0. (4.60)

We find

D

(ftftx

)=

Bt1(t1, t2)

2√B(t1, t2)2 − 1

((P 2 + 4Q+ 8Ptx + 8t2x)B

2(t, t1)−

−(P + 2tx)2 + 4(P + 2tx)

√Q+ tx(P + tx)B(t, t1)

√B2(t, t1)

) 12

,

ft1 = −1

2

(P + 2tx +

2√Q+ tx(P + tx)B(t, t1)√

B2(t, t1)− 1

)Bt1(t, t1).

Using the identities above, we check the coefficients of the terms,

(√Q+ tx(P + tx)

)i(tx

)jfor i, j = 0, 1, 2

60

in (4.60). Thus we obtain

(i, j) = (1, 1)⇒2B(t, t1)B

2t1(t, t1)√

B2(t, t1)− 1+

+2B(t, t1)

√B2(t, t1)− 1Bt1(t1, t2)

1−B2(t1, t2)= 0; (4.61)

(i, j) = (1, 0)⇒ P

2

(2B(t, t1)B2t1(t, t1)√

B2(t, t1)− 1+

+2B(t, t1)

√B2(t, t1)− 1Bt1(t1, t2)

1−B2(t1, t2)

)= 0; (4.62)

(i, j) = (0, 2)⇒B2(t, t1)B

2t1(t, t1)

1−B2(t, t1)−B2

t1(t, t1) +

+B2t1(t1, t2)

1−B2(t1, t2)−

2B2(t, t1)B2t1(t1, t2)

1−B2(t1, t2)= 0; (4.63)

(i, j) = (0, 1)⇒ P(B2(t, t1)B

2t1(t, t1)

1−B2(t, t1)−B2

t1(t, t1) +

+B2t1(t1, t2)

1−B2(t1, t2)−

2B2(t, t1)B2t1(t1, t2)

1−B2(t1, t2)

)= 0; (4.64)

(i, j) = (0, 0)⇒B2t1(t1, t2)

4(1−B2(t1, t2)).(P 2 − P 2B2(t, t1)− 4QB2(t, t1)

)+

+B2t1(t, t1)

( QB2(t, t1)

1−B2(t, t1)− P 2

4

)= 0. (4.65)

Then (4.61) and (4.62) are satisfied if

B2t1(t, t1)

B2(t, t1)− 1=

B2t1(t1, t2)

B2(t1, t2)− 1. (4.66)

Equality (4.63) is equivalent to

−( B2

t1(t, t1)

B2(t, t1)− 1−

B2t1(t1, t2)

B2(t, t1)− 1

)+

+2B2(t, t1)( B2

t1(t, t1)

B2(t, t1)− 1−

B2t1(t1, t2)

B2(t1, t2)− 1

)= 0. (4.67)

So equality (4.63) is satisfied. Note that (4.64) is a multiple of (4.63), so (4.64) is also

satisfied. Equality (4.65) is equivalent to

1

4

( B2t1(t, t1)

B2(t, t1)− 1−

B2t1(t1, t2)

B2(t, t1)− 1

)(P 2 − P 2B2(t, t1)− 4QB2(t, t1)

)= 0 (4.68)

and it is satisfied by (4.66). Hence all equations above are satisfied provided we have

equality (4.66). Equality (4.66) is equivalent to

Bt1(t, t1)

A(t, t1)= ±Bt1(t1, t2)

A(t1, t2).

61

ThenBt1(t, t1)

A(t, t1)does not depend on t and

Bt(t, t1)

A(t, t1)does not depend on t1. Hence we

find

Bt1(t, t1) = A(t, t1)ϕ(t1), (4.69)

Bt(t, t1) = ±A(t, t1)ϕ(t). (4.70)

Now we check the condition (4.12). We will show that B is a constant function. We

find

m = µ1tx(t2x + Ptx +Q) + µ2(t

2x + Ptx +Q)

32 + µ3(t

2x + Ptx +Q),

where

µ1 =2PB(Bt1 +Bt)

(P 2 − 4Q)(B2 − 1),

µ2 =2P√B2 − 1(Bt1 +Bt)

(P 2 − 4Q)(B2 − 1),

µ3 =(4Q− P 2 + P 2B)Bt1 + 4QBBt

(P 2 − 4Q)(B2 − 1).

which does not depend on t1. Hence, µ1, µ2 and µ3 do not also depend on t1. Using

(4.69) µ2 reduces to

µ2 =2P (ϕ(t1)± ϕ(t))

P 2 − 4Q.

Since µ2 does not depend on t1 we have either P = 0 or ϕ is a constant function.

Note that

µ1 =2PB(ϕ(t1)± ϕ(t))

(P 2 − 4Q)A,

and in both cases we find µ1 = 0 and µ2 = 0.

Firstly, assume φ = C to be constant. Using (4.69), µ3 reduces to

µ3 =C(4Q− P 2) + C(P 2 ± 4Q)B

(P 2 − 4Q)√B2 − 1

.

Derivative of µ3 with respect to t1 should be zero. Hence we get

0 =CBt1((P

2 ± 4Q) + (4Q− P 2)B)

(P 2 − 4Q)(B2 − 1)32

,

which gives Bt1 = 0 or ((P 2 ± 4Q) + (4Q− P 2)B) = 0. Both cases imply that B is

a constant.

Secondly, assume that P = 0 then using (4.69) again, µ3 reduces to

µ3 =4Q(Bt1 +BBt)

−4Q(B2 − 1)= −ϕ(t1)±Bϕ(t)√

B2 − 1.

62

Derivative of µ3 with respect to t1 must be zero. Hence differentiating the above

equality we have

0 =Bt1(Bϕ(t1)± ϕ(t))

(B2 − 1)32

,

which is possible if either Bt1 = 0 or B = ± ϕ(t)ϕ(t1)

. In the first case, if Bt1 = 0

then Bt = 0 by (4.69), so B is a constant. In the second case, if B = ± φ(t)φ(t1)

then

µ3 = ±√ϕ2(t)− ϕ2(t1). Since µ3 does not depend on t1, φ must be a constant and

hence B is a constant. As a result, in both cases we get B is a constant function and

using (4.59) we get the statement of the lemma. �

Using these lemmas we have the following theorem.

Theorem 71 Let f be a nonlinear function with respect to tx. The equation (4.32)

has four dimensional characteristic x-ring and two dimensional characteristic n-ring

if and only if

f =c1η(t)η(t1)

txor f =

c1ec2(t+t1)

tx + P− P,

where c1, c2, P are arbitrary constants and η is an arbitrary function of one variable,

or

f =√B2 − 1

√t2x + Ptx +Q+Btx +

P

2(B − 1),

where B, P and Q are arbitrary constants.

Proof: The proof of this lemma easily follows from Lemma 67, Lemma 68, Lemma 69

and Lemma 70. �

Examples

The functions f given in the Theorem 64 lead to the following examples.

Example 72 The equation

t1x =γ(t)

γ(t1)tx −

γ(t)

γ(t1)σ(t) + σ(t1),

where functions γ and σ satisfy the following relations

(γ(t)σ(t))′ = γ(t)√B1 +B2(γ(t)σ(t)), B1, B2 ∈ R,

63

is Darboux integrable with x-integral F = (L(t3)−L(t1))(L(t2)−L(t))(L(t3)−L(t2))(L(t2)−L(t1)) , where L(t) =∫ t

0γ(τ)dτ and n-integral I = γ(t)tx − σ(t).

Example 73 The equation

t1x =γ(t)

γ(t1)tx −

γ(t)

γ(t1)σ(t) + σ(t1),

where functions γ and σ satisfy the following relations

(γ(t)σ(t))′ = γ(t)√B1 +B2(γ(t)σ(t))2 B1, B2 ∈ R,

is Darboux integrable with x-integral F = (eL(t)−eL(t2))(eL(t1)−eL(t3))

(eL(t)−eL(t3))(eL(t1)−eL(t2)), where L(t) =∫ t

0γ(τ)dτ and n-integral I = γ(t)tx − σ(t).

The functions f given in the Theorem 71 lead to the following examples

Example 74 The equation

t1x =c1η(t)η(t1)

tx

is Darboux integrable with x-integral F =∫ t30η−1(τ)dτ −

∫ t10η−1(τ)dτ and n-

integral I =tx

c1η(t)+η(t)

tx.

Example 75 The equation

t1x =c1e

c2(t+t1)

tx + P− P

is Darboux integrable with x-integral F = e−c2t3+c2Px − e−c2t1+c2Px and n-integral

I =tx + P

c1ec2t+

ec2t

tx + P.

Example 76 The equation

t1x =√B2 − 1

√t2x + Ptx +Q+Btx + 0.5P (B − 1)

is Darboux integrable with x-integral

F = (−8B3 − 4B2 + 4B − 1)t+ (8B3 − 2B + 1)t1 + (−4B2 + 2B − 1)t2 + t3

and n-integral

I = (B −√B2 − 1)n(

√t2x + Ptx +Q+ tx + 0.5P ).

For each of these examples it can be checked that F is an x-integral and I is an

n-integral by direct calculations.

64

CHAPTER 5

CONCLUSIONS AND FUTURE WORKS

In this thesis we studied Darboux integrable semi-discrete equations of the form

t1x = f(t, t1, tx),∂f

∂tx6= 0. (5.1)

In Chapter 3, we considered semi-discrete hyperbolic chains of exponential type

ui1,x − tix = e∑a+iju

j1+

∑a−iju

j

i, j = 1, 2, . . . , N (5.2)

where A = (aij)N×N is a N × N matrix. For the equation (5.2) it was conjectured

that the chain (5.2) is Darboux integrable if and only if A is the Cartan matrix (see

[41]). We proved this conjecture for the case N = 2 but the other cases of N still

remains as conjecture and needs to be proved.

In Chapter 4, we classify Darboux integrable equations (5.1) with four dimensional

x-ring and two dimensional n-ring and some restrictions on the form of function f .

For future studies we may consider classification of Darboux integrable equation (5.1)

with no restrictions on dimension of characteristic rings and/or form of the function f .

65

66

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70

CURRICULUM VITAE

PERSONAL INFORMATION

Surname, Name: Bilen, Ergün

Nationality: Turkish (TC)

E-mail: ergun.bilen@metu.edu.tr

EDUCATION

Degree Institution Year of Graduation

M.S. Bilkent University, Mathematics 2012

B.S. Bilkent University, Mathematics 2010

High School Bolu Izzet Baysal Anadolu Lisesi 2005

PUBLICATIONS

1. N. Zheltukhina, K.Zheltukhin and E. Bilen. On a class of Darboux-integrable

semidiscrete equations. Advances in Difference Equations, 182, 2017.

2. K. Zheltukhin and E. Bilen. On Darboux integrable semi-discrete systems of

exponential type. arXiv:1712.03788, 2017.

71